university of nigeria university mathemati… · and complex numbers. volume 2 covers calculus...

University of Nigeria Virtual Library

Serial No ISBN 978- 175- 373- 0

Author 1 OYESANYA, M.O

Author 2 Author 3

Title Introductory University Mathematics 3

Keywords

Description Introductory University Mathematics 3

Category Physical Sciences

Publisher Africana Fep

Publication Date 2000

Signature

UNIVERSITY MATHEMATICS 0

. e d ~ r s and Elemenfa . . ., I t ''1

INTRODUCTORY UNIVERSITY

MATHEMATICS

Analytic Geometry, Vectors and Elementary Dynamics

J.C. Amazigo, I.A. Adjaero, C.E. Chidume, G.C. Chukwumah, A.K. Misra, A.D. Nwosu, C.A. Ntukogu, E.C. Obi, M.0. Oyesanya, ,./ C.O. Uche,

Ph.D. (Harvard), F.A.S. - Editor Ph.D. (U. Conn) Ph.D. (Ohio State) Ph.D. (Nigeria) Ph.D. (Kanpur) Ph.D. (R.P.I.) M.Sc. (De Paul) Ph.D. (Toledo) Ph.D. (Nigeria) Ph.D. (London)

Department of Mathematics University of Nigeria, Nsukka

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Preface

This third volume completes the set Introductory University Mathematics which we hope meets the "

need that we first observed in the preface of Volume I. Since much of that preface remains appropriate we reproduce it here.

"The three volume set Introductory University Mathematics is the result of a series of informal discussions over coffee by a number of us a t which we lamented the unavailability of good affordable mathematics texts for Nigerian tertiary institutions. After a while we decided to take the bull '

by the horns and contribute our little bit to the solution of the problem. The volume benefits from our combined experience of [over] two hundred years of teaching mathematics to students of mathematics, agriculture, engineering, the biological, environmental, management, physical and social sciences. The set covers the first year mathematics curriculum for Polytechnics and Colleges of Education and the National Universities Commission's (NUC) Minimum Academic Standards Mathematics Curriculum for Nigerian Universities. Volume 1 is devoted to Algebra, Trigonometry and Complex Numbers. Volume 2 covers Calculus while in Volume 3 Analytic Geometry, Vectors -

and Elementary Mechanics are treated. We have assumed that the reader is familiar with mathematics at the level of the Senior .

Secondary School Certificate or the General Certificate of Education O'level or their equivalent. On 1 most topics, however, we review the subject matter to this level before developing new concept^ and methodology. We neither emphasize the abstract nature of the underlying mathematical theory nor provide just a catalogue of formulas or techniques for solving large classes of mathematical problems. Rather, we follow an intermediate approach of motivating the introduction of new topics and results; leading the reader to discover new concepts and ideas and generalization of known results and assisting the reader in developing some mathematical intuition. This approach is based on our philosophy that if the text emphasizes the understanding of the underlying ideas and principles then the serious student would be in a position to tackle with confidence either problems arising from euch ideas in whatever guise they confront him or her or to appreciate further extensions of such principles and ideas."

Volume 3 contains seven chapters. The first four chapters cover analytic geometry. Chapter 1 is devoted to the straight line. The cirlce is presented in Chapter 2. The conic sections and their applications are covered in Chapters 3 and 4 respectively. Chapters 5 is devoted to vectors which provide the tools for the study of the third portion of the text dealing with elementary dynamics. In Chapter 6 motion of particles under prescribed acceleration (kinematics) is presented. Finally elements of Newtonian dynamics are discussed in Chapter 7.

The material is organised following the decimal system. Equations, examples, remarks and figures are numbered consecutively within each chapter. The latter two are numbered to indicate the chapter in which they occur. Thus, for example, Figure 3.17 is the seventeenth numbered figure in Chapter 3. Most sections are followed by exercises for the reader. The text contains a large number of carefully worked out examples that are used for further illustration of the ideas developed in the sections or subsections and several of these are not routine. Where possible some examples and exercises are formulated to reflect the Nigerian situation. We emphasize that the learning of mathematics cannot be a passive undertaking or something to be taken up only just before examinati6ns and any serious student must be prepared to work through the examples and tackle most of the exercises as part of the learning process.

We acknowledge the moral support and understanding of our families and thank our publishers for their patience with us.

April 1997

J.C. Amazigo Nsukka

Contents

Chapter 1 1 .o 1.1

1.2

1.3

1.4

The Straight Line Introduction Distances in the plane 1.1.1 Distances between two points in the plane 1.1.2 Internal and external divisions Area 1.2.1 Area of a triangle 1.2.2 Area of a polygon of n sides Equations of a straight line 1.3.1 The gradient of a line 1.3.2 Equation of a straight line 1.3.3 General equation of a straight line 1.3.4 Other forms of equation of a straight line 1.3.5 Sketching the graph of a straight line 1.3.6 Parametric equations of a s~a ight line Systems of straight lines 1.4.1 Angle between two straight lines 1 A.2 Parallel and perpendicular lines 1.4.3 Distance of a point from a line 1.4.4 Bisectors of angles between two non-parallel straight lines 1.4.5 Pencil of straight lines 1.4.6 Pairs of straight lines 1.4.7 Bisectors of angles between line pairs

Miscellaneous Exercises

Chapter 2 Coordinate Geometry of the Circle 2.1 General equation of a circle 2.2 Circles satisfying given conditions

2.2.1 Circle through three given points 2.2.2 Equation of a circle on a given diameter

2.3 Intersection of circle and line 2.3.1 Tangent to a circle 2.3.2 Normal to a point on a circle 2.3.3 Chord of contact

2.4 Systems of circles 2.4.1 Orthogonal circles 2.4.2 Radical axis of two circles 2.4.3 Coaxial circles

2.5 Equations of the form S + AL = 0; S + AS' = 0 Miscellaneous Exercises

Chapter 3 Conic Sections 3.1 Introduction 3.2 The Conic as a locus

3.2.1 Definition and examples 3.2.2 Standard forms of conics

3.3 Geometry and sketches of the conics 3.3.1 Translation of axes 3.3.2 The parabola 3.3.3 The ellipse

3.3.4 The hyperbola 3.3.5 Working in the original coordinates

3.4 Tangents and normals 3.4.1 Tangent and normal at a point on any curve 3.4.2 Tangent and normal at a point on a conic 3.4.3 Tangents from a point

3.5 Other forms of the equation of a conic 3.5.1 Parametric equations of the conic 3.5.2 Working with parametric equations


Chapter 4 Applications and General Equations of the Conics 4.1 Some applications of the conic

4.1.1 Parabola 4.1.2 Ellipse 4.1.3 Hyperbola

4.2 Polar equations of a conic 4.3 General equation of the second degree

4.3.1 The discriminant 4.3.2 Rotation of axes 4.3.3 Classification of general equation of second degree


Chapter 5 Vectors 5.0 W e e dimensional Cartesian coordinate systems 5.1 Definition and representation of vectors

5.1.1 Definition of vectors 5.1.2 Line segment representation of vectors 5.1.3 Cartesian coordinate representation

5.2 Algebra of vectors 5.2.1 Multiplication of a vector by a scalar 5.2.2 Addition of vectors 5.2.3 Scalar product of two vectors 5.2.4 Vector product of two vectors

5.3 Calculus of vector functions of a real variable 5.3.1 Differentiation of vector functions of a real variable 5.3.2 Integration of vector function of a real variable


Chapter 6 Kinematics of a Particle 6.1 Basic concepts 6.2 Components of velocity and acceleration in some coordinate systems 6.3 Motion under constant acceleration 6.4 Motion under variable acceleration

6.4.1 Rectilinear motion 6.4.2 Motion in a plane


Chapter 7 Dynamics of a Particle 7.1 Newton's laws of motion 7.2 Pulleys; Motion on smooth surfaces 7.3 Projectile motion 7.4 Simple harmonic motion 7.5 Impulsive motion

7.6 Motion of particles with variable mass Miscellaneous Exercises

Answers to Exercises Greek Alphabet Index

The Straight Line

1.0 Introduction

The cartesian coordinate plane has been introduced in Volume 1 of this series. In this chapter, we shall employ the method of coordinates to study the geometry of points and straight lines in a plane.

1.1 Distances in the Plane

1.1.1 Distance between two points in the plane

Let P ( x l , yl) and Q(x2, y2) be arbitrary points in the xy-plane and let the distance between the points P and Q be denoted by IPQ1 G d. See Figure 1.1.

FIG. 1.1

Let the perpendiculars from P and Q to the x-axis meet the axis at R and S respectively. Also let the perpendicular from P to Q S meet Q S at T. Then, clearly APTQ is a right-angled triangle. Moreover T has coordinates (xz, yl) and hence,

IQTI = ly2 - ~1 I (PTI = 1x2 - X I \ .

By Pythagoras' theorem applied to APTQ, we have

d2 = pq2 + )&q2 = (x2 - x ~ ) ~ + (y2 - yl)'

so that,

2 Thestraight Line

Since (22 - 2 1 ) ~ = (21 - ~ 2 ) ~ and (YZ - ~ 1 ) ~ = (yl - ya)', formula (1) can also equivalently be written as

Equation (1) gives a formula for calculating the distance between any two points (x i , yl) and ( z 2 , ~ 2 ) in the plane.

Example 1

Find the distance between the points A(-114) and B(-2, -3).

Solution

If d denotes this distance, then

so that, d = m = 5 . \ / 2 .

Example 2

Show that the triangle with vertices A(-l,2), B( l , -3) and C(3,2) is isosceles.

Solution

See Figure 1.2

FIG. 1.2

We compute: 1 ~ ~ 1 2 = [I - (-1)12 + [-3 - 21' = 4 + 25 = 29

1 ~ ~ 1 ~ = (3 + + (2 - 2)' = 16 + 0 = 16

pel2 = (3 - I ) ~ + (2 + 3)' = 4 + 25 = 29

so that (AB( = (BC(, and hence AABC is isosceles.

Example 3

Find the point equidistant from the points O(0, O), A(5, -5) and B(4,O).

Solut ion

Let the coordinates of the desired point be ( x , . ~ ) . Then the distance of the point (x, y) from O(0,O) 2 112 is d m ; from A(5, -5) is [(x - + (y + 5)2]1f2 and from B(4,O) is [(x - 4)2 + (y - 0) ] .

Since all the distances are equal we have:

(ii)

From (i), after squaring and simplifying, we obtain,

Similarly from (ii), 0 = -82 + 18.

These equations reduce to y - x + 5 = 0 and x = 2. Hence x = 2 and y = x - 5 = -3. The required point is (2, -3).

1.1.2 Internal and Ex te rna l Divisions

Let A(x1, yl) and B(x2, y2) be two points in the plane. We are inkrested here in the coordinates of a point P which divides the line segment AB in the ratio I : m, i.e., P such that A P : P B = I : m.

FIG. 1.3

Case 1 (Figure 1.3) The point P lies between A and B . (In this case we say that P divides AB internally in the ratio I : m). Let the coordinates of P be (x, y) and consider the triangles P A C and B P D . Then, lACl = )x - xl 1 , lPDl = 1x2 - X I , lPCl = Iy - yll and IBD1 = ly2 - yd . Clearly, the triangles P A C and B P D are similar. Hence

which yields

so that solving for x we obtain,

4 Thestraight Line

Similarly,

which yields

so that solving for y we obtain

Thus

Case 2 (Figure 1.4) Here P lies on A> produced. (In this case we say that P divides AB edernally in the ratio 1 : m).

FIG. 1.4

Construct the right-angled triangles PBC' and PAE'. Then, PC' = y - y2, PE' = y - yl, BC' = x - x2 and AE' = x - XI . Clearly triangles PBC' and PAE' are similar. Hence,

PB PC' --- P A - PE'

so that,

which yields ly - ly2 = my - my1 or,

i.e., (I - m)y = ly2 - my1 so that

BC' AE' 2 - 2 2 X - X I -- Also, - - 1

so that - - - 1

which yields, lx - 1x2 = mx - mxl or m m

1.1 Distances in the Plane 5

i.e., ( I - m)x = 1x2 - mx1, so that

( 5 )

Hence,

Remark 1.1

If P(x, y) is the midpoint of AB then m = I and so P(x, y )

Remark 1.2

In applying formulas (4) and (5) the following point is noteworthy for the correct application of the form. If P divides AB externally in the ratio 1 : m this can be considered as internal division if we replace 1 : m by I : (-m) as shown in Figur~ 1.5.

FIG. 1.5

Example 4

Find the points that divide into three equal parts a straight-line segment joining the two points A(5 , -2) and B(6,3) .

Solution

Let the desired points be C and D as shown in Figure 1.6,

FIG. 1.6

Then, since lACl = lCDl = IDBI, it follows that C divides AB internally in the ratio 1 : 2 (i.e.,

6 The Straight Line .

AC : C B = 1 : 2). Now, if C C(h, k) we have, using equations (2) and (3),

so that h = ?, k = -$. Hence

To calculate the coordinates of the point D we observe immediately that D divides AB internally in the ratio 2 : 1 (or D divides BA in the ratio 1:2). Regard D as dividing AB internally in the ratio 2 : 1, if D D(x, y) then,

Example 5

ABCD is a parallelogram with the coordinate of A, B and C given respectively by (0, -2), ( 8 , l ) and (4,8). Find the coordinates of D.

Solution

(See Figure 1.7). We shall solve this problem using the fact that the diagonals of a parallelogram bisect each other. A simpler solution may be obtained using the fact that the opposite sides are parallel.

FIG. 1.7

Let the coordinates of D be (h, k). Then, from A and C the coordinates of the midpoint of the diagonal AC is ( y, 9) = (2,3). This is also the midpoint of the diagonal DB. But the midpoint of D B is (y, y). Hence

h + 8 k + l - = 2 and - = 3

2 2 so that h = -4 and k = 5, and so D z D(h, k) = D(-4,5).

Recall that given a triangle ABC, the median of the triangle drawn from A is the line from A to the midpoint of the opposite side, BC. Similarly, the median from B is a line from B to the

1.1 Distances in the Plane 7

midpoint of AC; and finally the median from C is the line from C to the midpoint of AB. Recall also that the medians d a triangle are concurrent and meet at a point called the centroid. Furthermore, if AD is a median from A with M on AD as centroid t,hen lAMl : lMDl = 2 : 1. S w Figure 1.8.

FIG. 1.8

Example 6

Find the coordinates of the centroid of a triangle whose vertices are A(xl, yl), B(x2 , y2) and C(x3, 93). Solution

Let M denote the centroid of triangle ABC and let D denote the midpoint of B C as in Figure 1.8. Then

If M (2, y), since M divides AD internally in the ratio 2 : 1 we have

so that M = (2,y) = " + x2 + x3 , + y2 + y3 is the required centroid. 3

Exercises 1.1

1. Find the distance between the points A(-1, -2) and B(-4,3).

2 If A(-2,3) and B(4,2) are two fixed points and a point P(x, y) is chosen such that P A = P B , show that 122 - 2y - 7 = 0.

3 If C is the point (-5,3) and a point P(x, y) is chosen such that PC is a constant distance 2, prove that x2 + Y2 + 102 - 6y + 30 = 0.

4 Two vertices of an equilateral triangle are (1,O) and (a, 0). Obtain the possible coordinates of the third vertex.

5 Prove that the triangle with vertices R(O,6), S(-4,2) and T(-2,8) is a right-angled triangle.

8 The Straight Line

6 Let A(2, -I), B(1,-3) and P(z, y ) be such that PA = 2PB. Prove that 3x2 + 3y2 - 42 + 22y + 35 = 0.

7 Find the coordinates of the point which divides the line segment joining A(-!, 2 ) and ~ ( # , 3 ) externally in the ratio 3 : 2.

8 A point moves so that its distance from A(7,O) is equal to its distance from the y-axis. Prove that the equation of the locus of the point is given by y2 = 7(2x - 7).

1.2 Area

1.2.1 Area of a triangle

Consider the triangle ABC with vertices A(xl ,y I ) , B (x2 , y~ ) and C(x3,~3) . We want to find the area of triangle ABC. See Figure 1.9.

FIG. 1.9

Draw perpendiculars from A, B and C to meet the x-axis at P, Q and R, respectively. Then BQ = y2,

AP = yl and C R = y3. Also, QP = xl - 22, PR = 23 - X I and Q R = xs - 2 2 .

Area of triangle ABC = [Area of trapezium BQPA + Area of trapezium APRC - Area of trapezium BQRCl

For students who are familiar with determinants, the area of triangle ABC given in ( 6 ) can be represented as a 3 x 3 determinant given by

1 (7) Area of triangle ABC = absolute value of -

2 .

" 1 Yl 1 2 2 yz 1 x3 Y3 1

1.2 Area 9

The determinant in (7) is evaluated by the formula:

where, for arbitrary a , b , c, d, the determinant I : I is evaluated as

The expression on the right hand side (RHS) of equation (8) is obtained using the elements of the first row of the left hand side (LHS) as follows: The first term on the RHS of (8) is obtained by taking X I and multiplying it by the 2 x 2 determinant obtained from the LHS of (8) by deleting the row and column containing X I (see Figure 1.10a)

8 I

I 1 I

delete delete ' delete

a. FIG. 1.10

The second term is obtained by taking (-yl), (notice the minus sign introduced here), arid multiplying it by the 2 x 2 determinant obtained from the LHS of (8) by deleting the row and column containing yl (see Figure 1.10b). Finally, the third term is obtained by taking the 2 x 2 determinant obtained by deleting the row and column as shown in Figure 1 .10~ .

Remark 1.3

If the entries (i.e, the vertices of the triangle) in the 3 x 3 determinant in (7) are listed in an anticlockwise order then the determinant will have a positive value.

Example 7

Find the area of triangle ABC with vertices A(2,4), B(-1, -1) and C(4, 4).

Solut ion

Therefore, area of triangle ABC = +(?) = q.

10 The Straight Line

Example 8

Find the area of the triangle ABC with vertices A(7, -8), B(-2, -6) and C(1,5).

Solution

Therefore, area of triangle ABC = $ 1 - 1051 = F.

1.2.2 Area of a polygon of n sides

Given a polygon of n sides, to find the area of the polygon, it is always convenient to divide the polygon into several triangles with vertices as the vertices of the polygon and then find the areas of these triangles, take the sum of these areas to obtain the area of the polygon.

Example 9

Find the area of the quadrilateral ABCD with vertices A(2, O), B(-2, -2), C(-4, -4) and D(1, -7). Solution

A sketch of the quadrilateral is given in Figure 1.11.

t

FIG. 1.11

1.2 Area 11

So, area of quadrilateral ABCD = 21.

Area of quad. ABCD = area of AABC + area of ACDA

Exercises 1.2

Find the area of the triangle with vertices A(1,4), B(-2,8) and C(-7, -7).

Use the concept of area to show that the three points A(l , 0 ) , B(4, -4) and C(a, 1 ) are collinear.

The points P( l , -2), Q(6,lO) and R(26,25) are vertices of a parallelogram PQRS. Find (i) the coordinates of S , (ii) the area of the parallelogram.

Two vertices of triangle ABC are the points A(25,2), B(10, -10) and the centroid is the point (7,4). Find the coordinates of the third vertex C , and show that the triangle is right angled. Compute its area.

Prove that the quadrilateral PQRS given by P(-1, O ) , Q(3,2), R(4,5) and S(O,3) has opposite sides equal. Find the area of the quadrilateral.

Show that if t l , t2 and t3 are distinct then the points P(t:,t l) ,Q(ti , t2) and R(t:,tg) can never be collinear. Hint Write area as quadratic in t l .

Show that the three points ( 1 , O ) , (t:, 2t1) and (t;, 2i2) are collinear if tlt2 = -1.

Find the area of the quadrilateral ABCD where A, B,C, D have coordinates (3, -3) , ( 4 , 4 ) , (-4,3), (-2, -3) respectively.

If the vertices of a quadrilateral PQRS are given by ( 2 1 , yl), ( 2 2 , y2) , ( 23 , y3 ) and ( 2 4 , y4)

respectively, find the coordinates of the midpoint T of the line joining the midpoints of PQ and RS. Hefice, prove that the straight lines which join the midpoints of the opposite sides of a quadrilateral bisect each other.

A pentagon has vertices P(4,2), Q(-2,8), R(-6, -4) , S(1, - 11) and T(8, - 1 ) . Find the area of the pentagon. A hexagon has vertices A(0, 0 ) , B(-2,8), C(-5, -5) , D(0, - I ) , E(5. -8) and F(7, -1). Find the area of ABCDEF.

1 = - 2 0 1

-2 -2 1 -4 -4 1

+ - 1 - 4 - 4 1

1 -7 1 2 0 1

(vertices in anticlockwise order)

12 The Strejght Line

1.3 Equations of a straight line

1.3.1 The gradient of a line

Given a straight line AB, say, we shall define tbe slope or gradient of the line in terms of the angle 8 that it makes with the positive x-axis. See Figure 1.12. If this line is parallel to the x-axis the angle 0 is taken to be zero. I f i t is not parallel to the .].-axis, let it intersect the x-axis a t the point P.

(1)

FIG. 1.12

The straight line AB (Fig. 1.12) is said to make an angle 0 with the positive x-axis if an anticlockwise rotation of amount 8 of the x-axis about the point P and starting from P x rotates the x-axis to coincide, the first time, with AB. Thus 0 5 0 < r.

If the straight line makes an angle 0 with the positive x-axis, bhe tangent o f t h e angle 0, usually denoted by m, is called the gradient or slope of the line.

In general, the slope of the line joining two points A and B will be denoted by m A B .

Remark 1.4

(i) In Fig. 1.12(i), the angle 0 is acute, i.e., 0 < 8 < %, and in this case, tan0 is nonnegative, so that the slope is nonnegative. On the other hand, in Fig. 1.12(ii), 0 is obtuse, i.e., 5 < 0 < r, and so tan 0 is negative. Thus, slopes do not have to be necessarily positive.

(ii) If & line is parallel to the x-axis, then 0 = 0 and so tan 0 = 0. It follows then that if a line is parallel to the x-axis, its gradient is zero.

(iii) If a line is parallel to the y-axis, then 0 = + and since tan 5 = co, it follows that the gradient of a line parallel to the y-axis is co.

Now, suppose A ( x 1 , yl) and B(xz, y2) are two points on a given line. Draw through A and B lines parallel to the axes, and let these lines intersect a t C. See Figure 1.13.

I FIG. 1.13

1.3 Equations of a straight line 13

Then BC = xl - zz and CA = yl - yz. Furthermore, LABC = 8 so that,

and this is a formula for calculating the slope of a straight line (and hence the angle the line makes with the positive z-axis) when the coordinates of two points on the line are given.

Example 10

Calculate the slope of the line determined by the points A(l, -4) and B(-2, - 1).

Solution

Let ~ A B denote the slope of the line AB. Then

Example 11

Calculate the slope of each of the lines joining the pairs of points C(-3,6) and D(4,6), A(4,3) and B(4, -7) Solution The slope of the line C D is

so that C D is a horizontal line.

The slope of the line AB is

3 - (-7) 10 ~ A B = - - - and SO ~ A B = 0.

4 - 4 0'

This implies that the line AB is parallel to the y-axis. (i.e., AB is a vertical line.)

Example 12

Find the gradient of the line joining the points on the curve y = 3x2 - 23: + 1 whose x-coordinates are -1 and 2.

Solution Let the points be denoted by A(-1, yl) and B(2, y2) Then since A and B lie on the given curve, we must have

For A: yl = 3(-1)~ - 2(-1) + 1 = 6

For B: 92 = 3(2)2 - 2(2) + 1 = 9 9 - 6

Hence the points are A(-1,6) and B(2,9), so that ~ A B = = 1. 2 - (-1)

14 The Stra$:lt Line

Example 13

What are the gradiwts of the lines joining.the origin to the points of intersection of y = x2 and 2y = x + l ?

Solution

We first find the points of intersection of the two curves by solving the equations y = x2 and 2y = x + 1 simultaneously as follows:

2(x2) = x + 1 scl that 2x2 - x - 1 = 0

which implies 1

( 2 x + 1)(x - 1) = 0 so that x = -- or 1 2

Substituting in y = x2 yields y = (-i)2 and y = (1)' or y = f and y = 1 respectively. Thus, the points of intersection of the two curves are (-$, 4) and (1 , l ) . If rnl and m2 are the gradients of the lines joining the origin t o these points respectively, then

' - 0 m l = L - 1 -- 1 - 0

0 - 2 and m 2 = - = 1 .

-- - 2 1 - 0

1.3.2 E q u a t i o n of a s t r a i g h t line

Suppose a line goes through the point A(xl , yl) and makes an angle 0 with the positive x-axis. Let ..

P ( x , y) be any other point on the line. Let the parallel to Ox through A(xl , yl) meet the parallel . to Oy through P ( x , y) a t H, then L P A H = 0 and L A H P = 90'. See Figure 1.14.

FIG. 1.14

From Fig. 1.14, A H = x - x l , P H = y - yl so that , if 0 # 90°,


where m = tane. Equation (10) is the required equation of the line passing through the point A(xl, yl) and making an angle 0 # 90° with the positive x-axis. If 0 = 90°, then x = xl becomes the required equation.

Recall that m is the gradient (or slope) of the line found above. In forming the equation of any line, it suEces to know two facts about the line. These are:

(i) the coordinates of any one point on the line (ii) the slope (or gradient) of the line.

Suppose these two facts are known. Let the point on the line be A(xl , yl) and the slope of the line be m. Then, the equation of the line can be written down at once as (10). Thus, whenever we have to find the equation of a straight line we should be able to extract or compute from the data the coordinates of one point on the line and the slope of the line. Then we can immediately write down the equation of the line using equation (10).

Example 14

Find the equation of the line passing through the points A(-2,3) and B(4,2).

Solution

We can take e i ther A(-2,3) o r B(4,2) as our desired point on the line. Let us take A(-2,3). Now, we need to find the slope of the line. Since A and B lie on the line, its slope must be the same as that of the line segment joining A and B. If we denote this slope by m, then

We can now write down the required equation of the line using formula (10) as

which yields 6y - 18 = - x - 2 so that we obtain

as the requiqed equation of the straight line.

Example 15

Find the equation of the line passing through the points P(-1, -4) and Q(3, l) .

Solution

Taking Q(3 , l ) as a fixed point on the line, the equation of the line is given by

which simplifies to 5x - 4y = 11.


Example 16

Find the equation of the straight line which makes intercept of 3 on the y-axis and intercept of -2 on the x-axis.

Solution

FIG. 1.15

Since the line makes intercept of 3 on the y-axis then the coordinates of the point where the line crosses the y-axis has x-coordinate as zero. Similarly, since the line makes intercept of -2 on the z-axis, it must cross the x-axis at the point (-2,O). Let A(0,3) and B(-2,O) be the points of intersection of the line with the y- and x-axes respectively (Fig. 1.15). Then

so that the required equation of the line is

which simplifies to 32 - 2y + 6 = 0.

1.3.3 General equation of a straight line

Let us again consider the equation of the straight line of slope m which passes through the point ( X I , ~ l ) , i.ev

y - y1 = m(x - xl) .

We can rewrite this equation as y - m x = yl -mxl

where we observe that the right hand side, which is yl - mxl , is a constant. This equation can then be written more generally as

where p,q and r are constants. Equation (11) is called the general form of the equation of a straight line.


It can be shown that if px + qy + r = 0 represents the same straight line as p'x + qty + d = 0 then, their corresponding coefficients are proportional, i.e., for some constant k , we have

p = kpl, q = kg1 and r = kr'

Example 17

Find the equation of the straight line through A(-4 , -1) and B(2 , -3) and express it in the general form.

Solution

Hence, the equation of the line is 3

which simplifies to 3 y + 9 = - x + 2

and, in the general form, this is given by

Example 18

Find the values of I and m if l x + my + 5 = 0 represent the same straight line as 2y - 3x + 1 = 0.

Solution Since the' two equations represent the same straight line, then we must have,

so that I = 10 and m = -15.

1.3.4 Other forms of equation of a straight line

In section 1.3.3 we 'introduced the general form of the equation of a straight line. In this section we study other forms. We shall discuss the following forms:

a . The tangent form, b. The slope-intercept form, and c. The intercept form.

a. The tangent form The form of the equation of a straight line given in equation (10) of section 1.3.2 is usually called the tangent form. The constant m in that equation (which is the slope or gradient of the line) represente the tangent of the angle the line makes with the positive direction of the x-axis.


b. The slope-intercept fo rm Consider the equation of a straight line given in the tangent form as

Simplifying this equation we obtain

which yields, y = mx + (yl - mxl) or

where we have denoted the constant (yl - mxl) by c. Equation (12) is called the slope-intercept form of the equation of the straight line. Given the equation of any straight line, once y is made the subject as in equation (12), the coefficient of x gives the slope of the line. From equation (12), when x = 0 we have that y = c so that (0, c) lies on the line. Since all points on the y-axis must have x-coordinate zero the constant c gives the intercept on the y-axis.

Two special cases of equation (12) arise. We discuss these two cases below. Case I The required straight line passes through the origin. In this case the intercept on the y-axis is zero. So, in formula (12)) c = 0, and the equation of the line becomes

Case I1 The required line is parallel to the x-axis. In this case, the angle it makes with the positive direction of x-axis is zero, and since tan0 = 0, we have m = 0, so that equation (12) reduces

Example 19

Find the equation of the line passing through the points A(1,2) and B(-1,O).

Solut ion

The required equation is given by y = mx + c where m and c are to be determined. The constant m is easily determined from A( l , 2) and B(-1,O) as

Thus, the required equation is y = x + c, where c is to be determined. Since A(1,2) (or B(-1,O)) lies on the line, the coordinates (1,2) (or (-1,O)) must satisfy the equation of the line. Hence, using the condition that (1,2) must satisfy the equation, we have

2 = 1 + c so that c = 1.

Hence, the required equation of the line is y = x + 1.

Example 20

Find the equation of the line which makes an angle of 30° with the positive direction of the x-axis and passes through the point (-1, -2).


Solution

The required equation is given by y = mx + c, where m and c are to be determined. Now, m = tan30° = A, so that.the required equation is given by

But (-1, -2) lies on the required line, and so must satisfy the equation of the line. Hence,

Hence, the required equation of the line is

c. The intercept form Suppose a straight line makes an intercept of a on the x-axis and an intercept of b on the y-axis, a # 0, b # 0. Let A and B be the points where this line interesects the x- and y-axes respectively. Then the coordinates of A must be (a,O) and the coordinates of B must he l n h ) . Thus,

The equation of the line is then given by

so that, a y - ab = -bx or bx + a y = ab.

Dividing both sides of this equation by ab (since a # 0, b # 0) we obtain

Equation (13) is called the intercept form of the equation of a straight line. Observe that the intercept on the x-axis which is a is under x in equation (13) while the intercept on the y-axis which is b is under y.' The right hand side of the equation is 1.

Example 21

Find the equation of the straight line which makes intercept of -: on the x-axis and intercept of 7 on the y-axis.

Solution

By formula (13), the required equation is given by


which simplifies to

Example 22

Find where the straight line 32 + 4y = 7 intercepts the coordinate axes.

Solut ion 32 4y

32 + 4y = 7 implies (dividing by 7) - + - = 1 i.e., 7 7

and this implies the line intercepts the x-axis at (;,0) and the y-axis at (0,;).

1.3.5 Sketching the graph of a straight line

In order to sketch the graph of a straight line, it is sufficient to locate any two points on the line by means of coordinates. Two such points are easily located. Given the equation of a straight line, one can easily define the coordinates of the points of intersection of the line with the coordinate axes. On the x-axis, y = 0. So, putting y = 0 in the equation of the straight line we can find from the given equation the abscissa of the point of its intersection with the x-axis. On the y-axis, x = 0. Putting x = 0 in the equation of the strxight line we can find from the given equation the ordinate of the point of its intersection with the y-axis.

Example 23

Sketch the graph of the straight line defined by the equation 32 + 4y + 12 = 0.

Solut ion

If z = 0 then 4y + 12 = 0 so that y = -3. Thus the point (0, -3) lies on the line. Again, putting y = 0 we obtain. 32 + 12 = 0 eo that x = -4 and hence (-4,O) lies on this line. We now d o t the two points (0, -3, (-4,0) and join the two points by a straight line to obtain the graph of the line descr&&d by 32 + 4y + 12 ;0. See Figure 1.16.

FIG. 1.16 Equatioil of the line 3x + 4y + 12 = 0.


Example 24

Draw the graph of the straight line defined' by 32 - 2 y = 0.

Solution Put x = 0 to get -2y = 0 which implies y = 0. So, the point (0,O) lies on the line. If we now put y = 0, we still obtain that the point (0,O) lies on the line. One point is not sufficient to draw the graph of the line. So we assign any value different from zero to either x or y and compute the other. For example we may let y = 3 so that 32 = 2 y = 2(3) = 6 which yields x = 2. Hence (2,3) lies on the line. We have now obtained two distinct points on the line, i.e., (0,O) and (2,3). We plot thette points and join them by a &might line to obtain the graph of 32 - 2 y = 0. See Figure 1.17.

FIG. 1.17 The graph of the line 32 - 2 y = 0.

Example 25

Draw 'the graph of the straight lines represented by (i) x = -2, (ii) x = 7, (iii) y = 4 (iv) y-= -6 (v) x = 0 (vi) y = 0 .

Solution If the equation of the straight line is y = c or x = a then the construction of the line is reduced to drawing through the given point (0, c) or (a, 0) a straight line parallel to the x-axis or y-axis respectively. The graphs of the given equations are sketched in Figure 1.18.

, x = 0(he y-axis) , t

FIG. 1.18


1.3.6 P a r a m e t r i c aqua t ions of a s t ra igh t l ine

The straight line through the point (11, yl) whose gradient is tan I9 has the equation

y-y1 = (x-x1)tan19

x - X l -- ?T - '-" = t (say), B # o , - . cosD sin 0 2

Therefore,

Elimination of O yields t a = (x - x1)2 + (Y - y1)2 and so t represents the distance between a variable point (x , y) and the fixed point (x l , yl) on the line. Hence, different values o f t give different points on the line. To each point on the line corresponds a real value of t . The point (x l , yl) corresponds to t = 0. The equations

are called the parametric equatioils of t h t line and t is called the parameter. The equations

also represent a straight line through (xl , yl) since the elimination of t yields the linear equation

In this case, t in general, does not represent the distance between (x, y) and (xl, yl).

Example 26

Find the equation of the chord of the curve 3x2 + 4 ~ ' = 28 whose mid-point is (I., 1) and find also the length of this chord.

So lu t ion

FIG. 1.19


Let 6 be the angle the desired chord makes with the positive x-axis. Then t L x parametric equations of the line are

X = 1 + t ~ 0 ~ 6 , y = l + t s i n 0 . ti>

Where the line meets the given curve we have

3(1 + t cos 0)' + 4(1 + t sin 0)' = 28.

Multiplying the terms out and collecting like terms in t gives

(3cos20+4sin2fI) t2 +2(3cosfI+4sin6) t - 2 1 = 0. (ii)

Let the roots of this equation be t l > 0 and t2 . Since t = 0 is the midpoint of the chord, t l = -t2. Tha t is, t l + t2 = 0, or the sum of the roots of the quadratic

3 cos 0 + 4 sin 0 = 0.

Therefore tan 0 = -$. Equations (i) then implies

equation is zero. Hence

(iii)

from which we have 3 x + 4 y - 7 = 0

which is the required equation of the chord. The'length of the chord is 2tl . From equations (ii) and (iii)

t: = 21 - - 21 - - 21(3)

3 cos2 0 + 4 sin2 t9 3(9) sin2 0 + 4 sin2 0 28 sin2 8 or

3 t l = -

2 sin 6 4 15 15

But from (iii) sin0 = -, therefore t1 = - Hence the length of the chord is 2t1 = 7. 5 8 '

a

Example 27

A line is drawn through the fixed point P(cr,P) to cut the circle x2 + y2 = r 2 a t A and B. Prove that lPAl . IPBJ is independent of the gradient of the line.

Solution

FIG. 1.20

24 The Straight Liw

Let the parametric equations of the line PAB (me Figure 1.20) be

Where the line cuts the circle we have ( a + t cos B)2 + (P+t sin ~ 9 ) ~ = r2 SO, t2 + 2(Psin B + a cos 8) t + a2 + ,B2 - r2 = 0. If the roots of this equation are t l and t2 then from the theory of roots of quadratic equations, we obtain

t l t 2 = ] P A ] . lPBl = a2 + P2 - r2

which is independent of 8.

Exercises 1.3

Obtain the equation of the straight line joining the points

(iii) (at:, 2at 1), (at;, 2at (iv) (a cos el, b sin el), (a cos Ba, b sin 82)

Simplify your answer as much as possible in (iii) and (iv).

Find the equation of the straight line with intercepts 4 and -2 on the x-axis and y-axis respectively. Write down the following equations in the intercept form and hence write down the intercepts made by them on the axes.

(i) 3 x + 4 y + 7 = 0 (ii) y = m x + c , c # O , m # O

(iii) lx + t y + n = 0, ltn # 0

Find the equation of the straight line which cuts off a segment equal to 5 on the y-axis and makes an angle of 8 = 150" with the positive x-axis. Determine the slope and the y-intercept of the straight line represented by the equation 5x - 7y + 13 = 0. Find the equation of the straight line that has equal x- and y-intercepts and passes through the point (- 1,2).

Show that if px + qy + r = 0 represents the same straight line as p'x + q'y + r' = 0 then their corresponding coefficients are proportional, that is, for some constant k, p = p'k, q = q'k, r = r'k. The straight line x = t cos B - g, y = t sin B - f cuts the curve x2 + y2 + 292 + 2 f y + c = 0. Determine the values o f t at the points of intersection and show that they are independent of 8. What can you deduce about the curve from this result?

1 :I Systems of straight lines 25

1.4 Systems of straight lines

1.4.1 Angle be tween two straight lines

By the angle 0 between two lines AB and CD (see Figure 1.21) we shall mean the angle through which the straight line AB must be rotated in an anticlockwise direction until At) is parallel to C D or the angle through which the straight line CD must be rotated in an anticlockwise direction until CD is parallel to AB. For non-parallel and non-perpendicular lines there are two such angles, one acute and one obtuse and their sum is 180". If AB is originally para.lle1 to C D we shall take the angle between AB and C D to be zero.

FIG. 1.21

In our discussion below we shall assume, without loss of generality, that the two straight lines under consideration intersect at B (see Figure 1.22), so that we shall consider the angle between the lines AB and BC. In this setting, the angle between the straight lines BC and AB is taken to mean the angle 0 through which the straight line AB must be rotated in an anticlockwise direction about the point B until AB coincides with CB. Now, let the straight lines AB and CB be represented by '

the equations y = m l x + c l and y = m2x+ c2,

respectively. Let AB, CB make angles d l , 4 2 with the x-axis respectively as in Figure 1.22.

FIG. 1.22

From AACB, (b2 = 41 + 9 so that 0 = 42 - 41. This implies that

tan q ! ~ ~ - tan 41 tan 0 = tan(d2 - 41) =

1 + tan 42 tan 41 But t an& = slope of AB = ml , and tan42 = slope of CB = mz. Hence,


If the roles of the two lines are interchanged then

ml - m2 tan 0 =

1 + m2ml

Thus if we consider 0 as the a c d e angle between the two lines then

Example 28

Find the angle between the straight lines y = 4 2 - 7 and y = 3x + 1 .

Solution

From the equations we have ml = 4 , ma = 3 so that if 0 is the acute angle between the lines. then

and 0 = arctan (A) .

Example 29 ,

Find the equations of the straight lines which pass through the origin and form an angle of 60° with the straight line y = 22 + 1 .

Solution

From the data , 0 = 60'. The slope of the given straight line is 2. Let the slope of the required line be m. Then.

which upon squaring md siinplifying yields l l m 2 + 16m - 1 = 0. T h e solutions of this quadratic equation are

172 = -8 it 5,h 11 '

\\'e can iiow write down ,,lie equat,ioris of tlhe lines, using the fact that (0,O) lies on each line, as

-8 + 5 a -8 - 5 4 1, -o=

11 ( x - 0) and y - O =

11 ( x - 0)

-8 + 5& -8 - 5 f i y = 11

x and y = 11

x.

Example 30

Given the points A(-2,4), B(O,5) and C(-1, -1) determine the angles ABC and BAC.

1.4 Systems of straight lines 27

Solution

A sketch of AABC is p i v c , ~ ~ i l l Figure 1.23.

Y 4

cc-1.-ii 1 FIG. 1.23 1

Let = LA BC and Od4 = L BAG'. Then,

6 - 1 2 - 11 - 9 - ( - 5 ) 11 tanOB = - and t-anBA = - --

1 +6(;) 8 1 5 - 3

'Therefore, 11 11

LABC = dB = arctan (-) and L BAC = 0~ = arctan (--). 8 3

1.4.2 Parallel and perpendicular lines

If AB and Cl? in Subsection 1.4.1 are parallel then 0 = 0. Substituting 0 = 0 in (14) we have

and thus ma - ml = 0 or m l = ma. Tha t is, 2f two lznes are parallel lhey have the same slope. If A B and CU are perpendicular, the angle between them is 90° and tan 90" = CQ. Again from

(141,

00= m2 - ml

1 + m2m1

and this implies tha t the denominator of the right hand side must be zero, i.e., m2ml = -1. Thus , tf two strazght lzncs are perpend~cular, the product of their slopes is -1 (or the slope o f one is the nrgatzve of the reczproral of the slope of the other) .

a

Example 31

Find the equation of the line through (1 ,2) which is perpendicular to the line 4x + 3y + 12 = 0.

Solutiou

\5'e have one point (1.2) on the required line. If we can find the slope of the required line we can then write down its equation. Since the required line is perpendicular to the line 42 + 3y -t- 12 = 0,


its slope will be the negative of the reciprocal of the slope of 4x + 3 y + 12 = 0. To find the slope of this line we make y the subject of the equation to obtain

So, the slope of this line is -9 . Hence, if rn denotes the slope of the required line, we must have

As remarked above the equation of the required line can now be written down, since (1,2) lies on it , as

3 y - 2 = z ( x - 1 )

which simplifies to 3 x - 4 y + 5 = 0 .

Remark 1.5

A close examination of the given equation of the straight line in Exnmple 31 and equation (*) reveals that the coefficients of x and y in the given equation are interchanged in equation (*) but with the sign of the coefficient of y changed. The constant terms in the two equations are different. This observation is not peculiar to Example 31 above and can, in fact, be used to find the equation of a line perpendicular to a given line and passing through a given point.

In order to find the equation of a line perpendicular to the line ax + by + c = 0, a, b # 0, and passing through ( x l , y l ) let us denote the slope of the required line by m. Then from ax + b y + c = 0 we obtain.

so that the slope of the given line is (-f). Hence, the slope of the required line equals - l / ( - a / b ) = i. Since the required line passes through ( x l , y l ) we can now write down its equation as,

which simplifies to bx - a y + ( ay l - b x l ) = 0 .

Again, we observe that the coefficients of x and y in the given equation are interchanged in this equation but with the sign of the coefficient of y changed.

We give more examples.

Example 32

Find the equation of the line passing through (1, -2) and perpendicular to the line with equatian 3x + 2y - 7 = 0 .

Solution

The required equation is given by 22 - 3 y = k

(obtained by interchanging the coefficients of x and y in the given equation but changing the sign of the coefficient of y. It is also possible to retain the sign of the coefficient of y and change the sign of the coefficient of x). The quantity k is a constant. To determine it we use the fact that (1, -2) lies on the line. So, 2(1)'- 3(-2) = ic or k = 2 + 6 = 8. Hence, the required equation of the line is

Example 33

A perpendicular is drawn from P(1,2) to the line with equation 3x +4y - 12 = 0 and meets the line a t Q. Find tlw coordinates of Q (The point Q is called the foot of the perpendicular from P(1,2) t o the lane 32 + 4y - 12 = 0).

Solution

The line PQ is perpendicular to the given line. Equation 01' the line P Q is given by 42 - 3y = k. This line passes through the point P(1 ,2) . IIence,

Therefore, the equation of the line P Q is given by 42 - 3y = -2. To find the coordinates of Q we solve the equations

3 ~ + 4 y - 1 2 = 0

and 4 2 - 3 y + 2 = 0 , (9

(ii)

simultaneously. The solution of equations (i) and (ii) is x = and y = so that Q has coordinates

Example 34

Find the equation of the line through (1,2) which is parallel to the line 22 + 5y + 9 = 0.

Solution

The required line must have the same slope as the 22 + 5y + 9 = 0. To find the slope we make y the subject of this equation to obtain -

2 9 y = -- 5 x - 5

and the slope of this line is -$ . Since the required line is parallel to this line, it must also have the slope -$. The point (1,2) lies on the required line and so its equation is given by

which simplifies to give

30 T h e Straight Line

Remark 1.6

Equation (*) and the given equation in Example 34 have the same coefficients of x and y . The only difference between the two equations is l l ~ e constant term. This is not mere coincidence and can be used to quickly compute the equation of a line parallel t o a gzven l ine and passing through a given point. To derive this fact for the general equation ax + b y + c = 0, b # 0, is straight forward and is left as an exercise for the reader, i.e., the required equation always has t h e s a m e coefficeints of x and !/ as the given equation. T h w , to find this qua t ion , all that is required is t o write down the given equalion, replac 111% whatever constant is in it by a constant k, say. Then t o determine the value of k we use the fact that the given point lies on the required line. We illustratc: this below.

Example 35

Find the equation of the line passing through ( -2 ,3) and parallel to the line with eq~lation 5.r.+3y = ti.

Solution

The required equation is given by 52 + 3 y = b , where k is a constant to be determined. The point ( -2 ,3 ) lies on this line. So,

Thus, the required equation of the line is given by 52 + 3 y + 1 = 0.

Example 36

Find the coordinates of the mirror image of A(@,@) in the straight line ax + b y + c = 0, a # 0, b # 0.

Solution

Let B(cul, ,O') be the mirror image of A(@,@) in the line ax + b y + c = 0 and let A B meet the given line a t G (See Figure 1.19). If C, D are points on the given line,

Gradient of C D = -: (slope of the given line)

Gradient of A B = -I/(-$) = f .

Let 6 be the angle A B makes with the positive direction of x-axis. Then,

a tan 6 = -.

b

FIG. 1.24 Mirror image of A on ax + b y + c = 0


The parametric equations of AB are

(ii)

(iii)

so that for some t o

and the coordinates of G (which is the mid-point of AB) are given by

G lies on AB so that 1 1

a ( a + -to cos 8) + b(p + -to sin 8) + c = 0. 2 2

Solving for to we obtain

l o = -2(aa + bp + c) a cose + bsin O '

Hence the mirror image is at

obtained by substituting for tan 8 and cot 8 from (i).

1.4.3 Distance of a point from a line

Let a straight line be represented by px + qy + r = 0 and suppose P ( x l , yl) is a point not on the line. In this subsection we are interested in the line px + qy + T = 0. See Figure 1.25.

finding the perpendicular distance, dl from P ( x l , yl) to

FIG. 1.25

Let the foot of the perpendicular from P to the line px + qy + T = 0 be denoted by Q. To find the distance d(= J P Q J ) we should first find the coordinates of Q and then use the distance formula to compute the distance d. In order to find the coordinates of Q , however, we shall first find the equation of the line P Q and solve that equation simultaneously with px + qy + r = 0 to obtain the


coordinates of Q. We now proceed to do this. Since P Q is perpendicular t o p x + q y + r = 0 , its equation is given by

9 x - PY = k, where k is a constant to be determined. Now ( x l , y l ) lies on this line. So, 9x1 - pyl = k , and the equation of the line PQ is then given by q x - py - k = 0 , where k = 9x1 - p y l . To obtain the coordinates of Q, we shall solve the equations

p x + q y + r = o

and, q x - py - k = 0 (9

(ii)

simultaneously. Multiplying equation (i) by q and equation (ii) by p and subtracting gives y(pf+qf) = - ( q r + p k ) or

+ pk y = -- p2 + q2 '

Substitution of this in (ii) yields

so that,

Hence,

We can now compute d as follows:

pZx1 + q2x1 - 9k + P T ] + [ y i p 2 + ylq2 + q r + pk

p2 + q2

q2x1 + qpyi + pr

I y ip2 + ylq2 + q r + pqx l - p2yl

p2 + q2 p2 + q2 using k = el - pyl

I

so that

(15)

1.4 Systems of straigh t lines 33

Forxnula ( 15 ) gives the perpendicular distance d of the point P ( x l , y l ) from the line px + q y + r = 0 . Observe that the numerator of the right hand side of ( 15 ) is exactly the general form of the equation of the line under consideration but with x, y replaced by z l , yl which are the coordinates of the point P ( x l , y l ) whose perpendicular distance from the line is required. The denominator of the right hand side of ( 1 5 ) is the square root of the sum of the squares of the coefficients of z and y in p x + q y + r = O .

Example 37

Find the perpendicular distance of ( - 2 , 3 ) from the line 42 - 3 y = 8. Solution

We first rewrite the equation of the given line as 42 - 3 y -. 8 = 0 . If d denotes the perpendicular distance of the point ( - 2 , 3 ) from this line, then by ( 15 ) we have

( 4 ( -2 ) - 3 (3 ) - 8) 1 - 25) d = --- - JW a - 5 .

Example 38

Find the distance between the two parallel lines 12x + 5y - 7 = 0 and 122 + 5 y + 5 = 0 .

Solution

It suffices to take one point on any one of the lines and then find the perpendicular distance of this point from the other line. Let us choose an arbitrary point on 12x + 5 y + 5 = 0 . We can set x = 0 and obtain 5 y + 5 = 0 or y = -1 . Thus ( 0 , - 1 ) lies on this line. If d denotes the perpendicular distance of this point from 122 + 5 y - 7 = 0 then, by ( 1 4 ) )

l l 2 ( 0 ) + 5 ( - 1 ) - 7 1 12 12 d = - -- -

, ~ T F - m - i 3 + Thus, the distance between the two parallel lines is g.

1.4.4 Bisectors of angles between two non-parallel straight lines

Let el a l x + b l y + cl = 0 and t2 E a2x + b2y + c2 = 0 represent two non-parallel straight lines. Then a l b 2 # a z b l . See Figure 1.26. We are interested in computing the equations of the lines which are the bisectors of the angles between t l and denote these bisectors as B1 and B 2 (see rig.

t2 . In general, there are two such bisectors. We shall 1.26).

*

4 , I ,

FIG. 1.26

34 The Straight ~ i n e

We observe immediately that any point on the bisector of the angle between the lines e l and & is equidistant from el and t2. Thus, if P(x, y ) is any point on this bisector, then the distance, of P ( z , y) from alx + bly ,+ cl = 0 equals its distance from a22 + b 2 y + c2 = 0. Hence, the equations of the two bisectors are given by

These equations are equivalent to the two equations

and

Collecting like terms in each equation gives

and

Observe that these equations are the equation of straight lines provided both coefficients of x and y do not vanish. (Show that these coefficients do not vanish since a2bl # alb2). Thus, these are the equations of straight lines whose slopes rnl and m2 are given respectively by

ml = - *-& and m2 = - ?*'* $+-A .*+he Observe that

mlm2 =

so that the two bisectors are always perpendicular to each other.

Example 39

Determine the equations of the bisectors of the angle between the lines x+2y - 1 = 0 and 2x+ y + 3 = 0.


Solution Let P ( x , y) be a point on a bisector of the angle between the lines x + 2y - 1 = 0 and 2x + y + 3 = 0 . Then, since this point is equidistant from the two lines

That is x + 2 y - 1 = 2 x + y + 3 and + + 2 y - 1 = - ( 2 ~ + ~ + 3 )

or x - y + 4 = 0 and 3 x + 3 y + 2 = 0 .

Thus, the equations of the bisectors of the angles between the lines x + 2 y - 1 = 0 and 2 2 + y + 3 = 0 are

2 x - y + 4 = 0 and X + ~ + ~ = O .

1.4.5 Pencil of straight l ines

A system of straight lines passing through a given point is called a pencil of slruighZ lines (Fig. 1.27) and the given point is called the vertex of the pencil.

FIG. 1.27

Consider the two distinct non-parallel straight lines

and set up the equation u kle l + k2e2 = 0

where kl and k2 are not both zero. The equation is linear in x and y and so represents a straight line. Furthermore, u vanishes at the point of intersection of the lines given by el = e2 = 0 . Thus, the equation u = 0 represents a straight line passing through the intersection of el = 0 and e2 = 0. Conversely, every straight line of the pencil determined by el = 0 and Z2 = 0 can be represented by the equation cle l + c2e2 = 0. Furthermore,


is a member of the pencil if a number ks exists such that kill + k212 + k313 = 0 . 1

Example 40

Obtain the equation of the straight line through the point of intersection of x + 3 y + 2 = 0 , x - 2 y - 4 = 0 and perpendicular to 51: + 2 y - 9 = 0.

Solution

Any line through the intersection of x + 3 y + 2 = 0 and x - 2 y - 4 = 0 can be written in the form

where kl and k2 are constants (not both zero) to be determined. This equation can be re-written as

This line is perpendicular to 5x + 2 y - 9 = 0 if the product of their slopes is - 1 , i.e., if

i t . , 5 ( k l + k2) + 2(3k l - 2k2) = l l k l + k;! = 0 or k2 = - I l k l . Substituting for k2 in equation (i) gives

x + 3 y + 2 - l l ( x - 2 y - 4 ) = 0 or 1 0 2 - 2 5 y - 4 6 = 0 .

Thus the required equation is 103: - 25y - 46 = 0 .

Example 41

Prove that the altitudes of a triangle are concurrent.

Solution

Let the vertices of the triangle be P l ( x l , y l ) , P2(x2 , y 2 ) and P3(x3, y3) (see Figure 1.28) .

FIG. 1.28

Then slope of P2P3 = ( y 2 - y 3 ) / ( x 2 - 23) . Equation of altitude through P l ( x 1 , y l ) is given b y


Similarly, equation of the altitude through P2(x2, y2) is given by

e2 (Z - X Z ) ( X ~ - 21) + (y - YZ)(Y~ - ~ 1 ) = 0. (ii)

Also, equation of the altitude through P3(x3] y3) is given by

e3 ( X - X ~ ) ( X I - 22) + (y - YS)(YI - ~ 2 ) = 0. (iii)

Now, since the sum of the three equations (i), (ii) and (iii) is zero,

klel + kzl2 + kal3 = 0,

for kl = kg = kg and so e3 is a pencil of el and 12. Thus, the three altitudes are concurrent.

1.4.6 Pairs of straight lines

The general homogeneous equation of the second degree in x and y can be written as

If a # 0, completing the square of this equation in terms of x we obtain, after first multiplying by a ,

(ax + hy)' - (h2 - ab)y2 = 0,

provided h2 - ab 2'0. So, Equation (i) represents the two straight lines

a x + ( h + d = ) y = ~

and ax + (h - d E ) y = 0.

These two straight lines clearly pass through the origin.

If a = 0, b # 0, Equation (i) may be written as

giving the two equations y = 0 and 2hx+ by = 0

which represent two straight lines passing through the origin.

If a = 0 = b, h # 0, then Equation (i) may be written as

(ii)

(iii)

which yields x = 0, y = 0 as the equations of two straight lines passing through the origin.


If b = 0, a # 0, the Equation (i) may be written as

and represents the lines x = 0 and ax + 2hy = 0.

Thus, Equation (i) always represents a pair of straight lines through the origin. Throughout this subsection, we shall assume b # 0. It is clear from Equations (ii) and (iii) that the lines represented by (i) are real and distinct if h2 > ab, real and coincident if h2 = ab. For h2 < ab the equations (ii) and (iii) have no real solutions. For b # 0 Equation (i) may be expressed as

Equating coefficients we obtain

Thus the equation ax2 + 2hxy + by2 = 0, b # 0, h2 2 ab represents a pair of straight lines y = mlx and y = m2x whose slopes ml and m2 are related by Equation (iv).

1.4.7 Bisectors of angles between line pairs

The line pa i r ax2 + 2hxy + by2 = 0, h2 > ab

If the equation ax2 + 2hxy + by2 = 0 represents the lines y = mix, y = m2x, then as noted in the last subsection ml + m2 = -? and mlm2 = f . We wish to obtain the equations of the straight line bisectors of the angle between this line pair.

g e t el be the line y = mlx and t2 be the line y = m2x. Furthermore, let el make an angle 81 with the positive x-axis and 1 2 make an angle 82 with the positive x-axis (Figure 1.29). Then tane l = rnl,tanO7.= ma.

I ' \ FIG. 1.29

If 0 is the angle made by the bisector of the angle between el and L2 then 0 - el = 82 - 8 SO that 28 = 81 + 82. Equations of the bisectors are given by

lr y = x t a n 8 , and y = x t a n ( e + - ) 2

or, y -x t anB=O, and y + ~ ~ ~ t e = 0


which imply

But 2

c o t e - t a n 0 = - tan 20

- - 2 since 28 = el + 82

tan(01+ 82)

- 2(1 - tan el tan 02) -

tan 01 + t'an 02

Thus the equation of the bisectors is y2 + q x y - a:? = 0.

Example 42

Write down the equation of the line pair bisecting the angles between the line pair 4x2-xy-3y2 = 0.

Solution

The equation of the desired line pair is given by

a - b Y +- h

xy - x2 = 0

where a = 4, b = -3,2h = -1, so that

is the required equation.

Example 43

Show that x - y = 0 bisects the angle between the lines 4x2 - l l x y + 4y2 = 0 and write down the equation of the other bisector.

Solution

The equations of the bisectors of the angle between the lines 4x2 - l l x y + 4y2 = 0 is given by

where a = 4, b = 4 ,2h = -11. Hence, the required equation becomes y2-x2 = 0 or (x-y)(x+y) = 0. Thus, x - y = 0 is a bisector of the angle between the lines 4x2 - l l x y + 4y2 = 0, and the equation of the other bisector is x + y = 0.


Two straight lines i n general

The standard form of the general equation of the second degree is

If this equation is to represent a line pair, the left hand side must be factorizable into two linear factors. Writing (i) as a quadratic equation in y, and solving, we obtain,

Now, y can be expressed in the form a x + p if, and only if, (bx + k)' - c(ax2 + 2hx + d) is a perfect square, i.e., if, and only if,

(bk - ch)' = (b2 - ac)(k2 - cd)

or, on simplification, c(acd + Phkb - ch2 - ak2 - db2) = 0

which yields, since c # 0, acd + Phkb - ch2 - ak2 - db2 = 0. (ii)

Now, the condition (ii) was derived on the assumption that c # 0. Suppose now that c = 0 and a f 0, we still obtain condition (ii) by setting c = 0 in Equation (i) and solving for x.

If a = 0 = c and b # 0, equation (i) becomes 2bxy+2hx+2ky+d = 0 or b ~ ( 2 ~ + ~ ) + k ( 2 ~ + $ ) = 0. The left hand side has linear factors only if = $, i.e., if d = q. This is the condition (ii) with a = 0 = c , so that in all cases (ii) is the condition for Equation (i) to represent a line pair.

This is the required condition for Equation (i) to represent a line pair. The condition (ii) can be put in determinant form as

Example 44

Prove that x2 + 6xy + 9y2 + 42 + 12y - 5 = 0 represents a pair of straight lines.

Solut ion

Comparing the given equation with ax2 + 2bxy + cy2 + 2hx + 2ky + d = 0 we identify the following:

a b h b c k h k d

a = l , 2 b = 6 , ~ = 9 , 2 h = 4 , 2 k = 1 2 , d = - 5 so tha t a = l , b = 3 , ~ = 9 , h = 2 , k = 6 a n d d = - 5 .

= O . (iii)

Now,

since the first two rows are proportional. Hence, the given equation represents a line pair. Alternative method: The given equation can be expressed as x2 + (6y + 4 ) r + gy2 + 12y - 5 = 0. Solving this equation for x gives

x = -(3y + 2) f ,,43y + 2)2 - 9y2 - 12y + 5

a b h b h k d

c k = 3 9 1 3 2

2 6 - 5 6 = 0 ,


This implies that x + 3 y - 5 and x + 3 y + 1 are factors of x2 + ( 6 y + 4 ) x + 9y2 + 12y - 5 . That is, x2 + ( 6 y + 4)x + g y 2 + 12y - 5 = ( x + 3 y - 5 ) ( x + 3 y + 1) = 0 and hence the given equation represents the pair of lines x + 3 y - 5 = 0 and x + 3 y . + 1 = 0 .

Example 45

Show that 2x2 + 3 x y - 2y2 + 52 - 10y - 12 = 0 represents two perpendicular straight lines and find their point of intersection.

Solution

To factorize this equation into the product of two linear factors we start by first factoring as follows:

We can now write

where a and /3 are constants to be determined. Expanding the right hand side we obtain

Equating coefficients we obtain c u + 2 p = 5 2cu - p = -10.

Solving we have a = - 3 , /3 = 4 . Hence,

and so'the given equation represents the two straight lines whose equations are

2 2 - y - 3 = 0 and x + 2 y + 4 = 0

or 1

y = 2 x - 3 and y = - - x - 2 2

with ml = 2 and m2 = -a, where ml and ma are the slopes of the above lines respectively. Since,

the two lines are perpendicular. To find the point of intersection of the lines, we solve the equations (i) of the lines simultaneously to obtain (g, -9) as the point of intersection.


Exercises 1.4

Find the angle between the lines 3x + 3y - 7 = 0 and 4x - 3y + 8 = 0.

Consider the equation ax + by + c = 0. Suppose b = 0 , a # 0 . Show that the equation of the line perpendicular to this line is given by ay = k where k is an arbitrary constant to be determined. If this line passes through ( x l , y l ) , show that its equation is y = yl.

a. Find the equation of the line passing through ( -1 , -1) and perpendicular to the line x + y = l .

b. Find the equation of the line passing through ( 1 , 2 ) and parallel to the line x + y + 3 = 0 .

a. Find the equation of the line passing through the intersection of the lines 22 - 3y +7 = 0 and x + y + 1 = 0 and parallel to the line 5x + 7 y - 8 = 0 .

b. Find the equation of the line passing through the intersection of the lines 22 - 3y + 7 = 0 and x + y + 1 = 0 and perpendicular to the line 52 + 12y = 13.

Let A(-1, I ) , B ( 2 , 5 ) and C(-2,8) be the vertices of a triangle. Find (i) angle ABC, (ii) angle BAC. Show that if the lines l I : a l x + bly + cl = 0 and 12 : a2z + b2y + c2 = 0 are parallel then alb2 = azb l , and if they are perpendicular then ala2 + blb2 = 0 . Use the method of this section to solve problems 7-12.

Obtaih the equation of the straight line which passes through the point of intersection of the lines x + y = 3 and 22 = y + 5 and

(i) passes through the origin (ii) is parallel to the line 5x - y = 4

(iii) is perpendicular to the line 7% - 5y + 2 = 0 (iv) makes angle 45O with the positive direction of the x-axis.

Show that the medians of a triangle are concurrent.

.Prove that the three perpendicular bisectors of the sides of a triangle are concurrent.

,Prove that the straight line ( 2 + t ) x + ( 1 + 2)y = 5 + 72 always passes through a fixed point whatever the value o f t and find the coordinates of this point.

Find k such that the straight lines x - 2y - 6 = 0 , 32 + y - 4 = 0 and k x + 4y + k 2 = 0 are concurrent. Obtain the equation of the line through the point of intersection of x + 6y - 7 = 0 and 32 - 2y + 2 = 0 and perpendicular to one of them. Find the equations of the lines represented by the following pairs of lines

(i) 4x2 - y2 = 0 (ii) 2x2 - 5xy - 3y2 = 0

(iii) -x2 + y2 + x + 3y - 2 = 0 (iv) 3x2 + x y - 2y2 - 182 + 17y - 21 = 0.

Form the equations which represent the following pairs of lines. (i) y = 0 , x = 4y

(ii) 3 2 - y = O , x + 3 y = O (iii) y = n x , y = m x .

Calculate the angle between the two straight lines given by x2 + 2xy - 4y2 = 0 .

Show that x2 + x y - g y 2 - x - 8y - 2 = 0 represents a line pair and calculate their angle of intersection. Prove that the line pair x2 + 4 x y + y2 = 0 and the straight line y = x + k form an equilateral triangle for all k .

1.4 Systems d ~ t r a i g h t lines 43

18 Show that the two line pairs lox2 + 8xy + y2 = 0 and 5x2 + 12xy + 6~~ = 0 contain the same angle.

19 a. Prove that the area of the triangle whose vertices are the points (0, O), (xl, yl), (x2, y2) is 3 1 ~ 1 ~ 2 - 22~11.

b. If 0 is the origin, and if the line 1x +my = 1 meets the pair of lines whose joint equation is ax2 + 2hxy + by2 = 0 in P (x l , yl) and Q(x2, y2), prove that the area of the triangle O P Q is

d m )am2 - 2hlm + bP1'

20 From a point Q(p,q) perpendiculars QM,QN are drawn to the straight lines given by the equation ax2 + 2hxy + by2 = 0. Prove that if 0 is the origin of coordinates, the area of the triangle O M N is

(aq2 - 2hpq + bq2)(h2 - ab)lI2 (a - b)2 + 4h2


1 Prove that the triangle with vertices A(-1, I), B(3,3) and C ( l , -1) is isosceles

2 Find the coordinates of the point which divides the straight line segment joining the two points A(-1,l) and B(3,3) internally in the ratio 3 : 1.

3 Two vertices of an equilateral triangle are (0,2) and (0,6). Obtain the possible coordinates of the third vertex.

4 Let A(l , 2), B(-1, 1) and P(x, y) be such that P A = k P B where k is some constant. Find the relationship between x and y. What is the nature of the locus of P when k = l ? Simplify your relationship for this case.

5 Given that the triangle ABC with vertices A(-1, -I), B(3,2) and C(a, -2) is right angled find all possible values of a.

6 A point P(x, y) moves so that its distance from A(O,3) is equal to its distance from the x-axis. Find the relation between x and y.

7 Find the area of the triangle with vertices A(-1, l ) , B(3,3) and C(1, -1).

9 Find the area of the quadrilateral with vertices (2, I), (-2,3), (-3, -1) and (1, -2).

10 a. Find the equation of the line passing through (-1,2) and perpendicular to the line x + 2y = 3.

b. Find the equation of the line passing through (2 , l ) and parallel to the line 22 + y = 1.

11 Find the angles of the triangle with vertices A(-1,2), B(0, -1) and C(3 , l ) .

12 Obtain the equation of the straight line which passes through the point of intersection of the l i n e s 3 x + y + l = O a n d y = 3 x + l a n d

(i) is parallel to the straight line 4x - y = 7 , (ii) passes through (1, l ) ,

(iii) is perpendicular to the straight line 42 + y = 5, (iv) makes angle of 30° with the positive y-axis.

13 Show that the bisectors of the angles of a triangle are concurrent.


14 Find the equations of the lines represented by the following pairs of lines: ( i ) x y + 3x - 2y = 6 , , .

(ii) x2 - y2 + 4 x + 4 = 0, (iii) 2x2 + 3xy - 2y2 - x + 3y - 1 = 0 .

15 Find the angle between the two straight lines given by 18x2 + 18y2 + 45xy + 9y - 2 = 0.

16 Prove that y2 - 4xy + x2 - 10y + 8x + 13 = 0 represents a pair of straight lines. Find their point of intersection and the angle between them.

17 If the lines ax2 + 2hxy + by2 = 0 meet the line qx + py = pq in points which are equidistant from the origin, prove that h(p2 - q2) = pq(b - a ) .

18 Show that the equation 3x2-4xy-4y2+14x+12y-5 = 0 represents twostraight lines, and find the combined equation of the bisectors of the angle between them. Hint Set x = 8-1, y = 1+2.

19 Prove that x2 + y2 + 2xy sin cr + 2ax cos cr + a2 = 0, for all values of cr # f 5, represents a pair of straight lines and show that the locus of their point of intersection is the curve x2 - y2 = a2. Hint: A2 + B2 = 0 c4 A =i 0 and B = 0.

Coordinate Geometry of the Circle 2.1 General equation of a circle

If a point moves, subject to certain restrictions, the path it traces out is called its locus. The locus of a point which moves such that it is always the same distance from a fixed point is called a CIRCLE. The fixed point is called the centre of the circle and the constant distance its radius. Our interest in this chapter is the study of the properties of circles.

Suppose the centre of a circle has coordinates (a, b) and the circle has radius r . Let (x, y) be the coordinates of an arbitrary point on the circle. Since the distance of (2, y) from the centre of the circle must be equal to r , using the formula (1) of Chapter 1 for the distance between two points, we have,

Equation (1) is the standard form of the equation of a circle with centre (a, b), radius r . I t gives both the coordinates of the centre and the radius of the circle. As an immediate consequence of this, if (a, b), the centre of the circle is the origin, its equation becomes

Expanding the standard form, we obtain

This last form is usually written as

where h = -a, k = -b and d = h2 + k2 - r2 . If the equation of the circle is given in this form, then by completing squares, we obtain

so that by (1) the coordinates of the centre are (-h, -k) and the radius is given by

r = d-, provided h2 + k2 - d 2 0.

Example 1

Find the centre and radius of the circle whose equation is given by 3x2 + 3y2 - 6x + 9y - 1 = 0.

46 Coordinate Geometry of the Circle

Solution

We first divide the given equation by 3. Thus, we have,

Comparing this equation with equation (3), we see that, 2h = -2,2k = 3 and d = -; so that h = -1, k = %. Thus, the centre of the circle is given by (-h, -k) = ( I , -%), and.the radius by

r = ,/h2 + k2 - d = J(-l)2 + (-$)2 - (-4) = fi. Aliter

We could complete squares to get

and deduce immediately that the centre is the point (1, - q ) and the radius r is fi, as before.

Example 2

Does the equation 5x2 + 5y2 - 102 + 69 + 25 = 0 represent a circle? Justify your answer.

Solution

We complete squares in the given equation. Before we do this, however, we must first make the coefficients of x2 and y2 unity. Thus, we obtain, .

which yields, on completing squares,

which does not represent a circle since r = fl is not a real number.

Aliter

We could compare the given equation with the general equation (3). Before we do this, we must again make the coefficients of x2 and y2 unity to obtain,

and comparing this equation with (3) yields h = -1, k = -: and d = 5 and this yields

so that the condition h2 + k2 - d > 0 is not satisfied. Hence, the given equation does not represent a circle.

2.2 Circles satisfying given conditions 47

Exercises 2.1

1 Show that each of the following equations represents a circle. Find its centre and radius. (i) 4x2 + 4y2 - 32 - 8y - 1 = 0

(ii) x2 + y2 - 42 - 6y + 9 = 0 (iii) 5x2 + 5y2 - lox + 6y - 1 = 0 (iv) 3x2 + 3y2 + 6x - 27y = 2 (v) z2 - 22 = 4y- y2,

2 Does each of the following equations represent a circle? Justify your answer. (i) x 2 + y 2 + x - y + 3 = 0

(ii) 2x2 + 3y2 + 42 - 6y + 3 = 0 (iii) (x - = 2 + (y + 3)2 (iv) x2 + y2 - x + y = 8.

2.2 Circles satisfying given conditions

2.2.1 Circle t h r o u g h t h r e e given poin ts

Given three points in a plane, there may or may not be a circle that passes through them. If the points are not collinear then such a circle exists. The equation of the circle may be obtained using any of the following properties of the circle as shown in the example below.

(i) The three points must satisfy the general equation (3) for some values of h , k and d. (ii) The centre of the circle is equidistant from the three points.

(iii) The desired circle is the circumcircle of the triangle with the the given points as vertices. If the points are collinear then no such circle exists (suggest why not).

Example 3

Obtain the equation of the circle through the three points A(2,6), B(4, -2) and C(-2,2).

Solut ion

Method 1

We may start by writing the general equation of the circle as

If we can find the values of h, k and d then we know the required equation of the circle. If we could obtain three simultaneous equations in the three unknowns then we would be able t o find their values. We may use the fact that since the three given points lie on the required circle, the coordinates of these points must satisfy the equation of the circle. In particular, the point (2,6) lies on the circle implies, 22 + 62 + 2h(2) + 2k(6) + d = 0 or

Similarly, using the fact that (4, -2) and (-2,2) lie on the required circle we have the following two equations:

8 h - 4 k + d = - 2 0 (ii)

and. - .4h + 4k + d = -8. (iii)


We now solve the three equations (i), (ii) and (iii) simultaneously to obtain h, k and d. For example, to get rid of d we observe that (i) - (ii) : -4h + lt3k = -20 @I (ii) - (iii) : 12h - 8k = -12 To eliminate h from (iv) and (v) we have:

(v)

(iv) x 3 : (v) is:

Adding: 9

40k=-72, so k = - - 5'

Substitution in (iv) yields:

Substitute in (iii) to get

- 4 ( - F ) + 4 ( - i ) + d = - 8

so that

Therefore, the required equation of the circle is

or 5x2 + 5y2 - 222 - 18$ - 48 = 0.

Me-lhod 2

Let K(-h, -k) denote the centre of the required circle. In this method we use the fact that the centre is equidistant from the points A, B and C. Now

Thus -h+4k = -5

Also

Thus h + k = - 4 (vii)

2.2 Circles satisfying given conditions 49

Solving equations (vi) and (vii) simultaneously we obtain, by adding both equations,

Substitution in (vii) yields h = - y. Hence the centre of the required circle is (-h, - k ) = ( F , g). We now use this centre with A, B or C to compute the radius of the circle. If r denotes the radius of this circle, using the point A, we obtain,

If now P(x , y) is an arbitrary point on the circle, we have that the distance of P from li' is r2, i.e., IPIr'I2 = r2 or

( x - ;)'+ (Y- = (2- F ) ~ + (6- :)2

which, on simplification, yields the equation of the required circle as

Method 3

For a third method of finding the equation of a circle passing through three given points, see Exercises 2.2, problem 9.

2.2.2 Equation of a circle on a given diameter

Given the end-points PI, Pz of a diameter of a circle we can determine the equation of the circle by observing that

(i) the centre of the circle is the mid-point of PIP2 and the radius is 3 1 ~ ~ ~ ~ 1 or (ii) if P is any point on the circumference then PlPP2 is a right angle (the diameter subtends a

right angle at the centre) as illustrated in the following examples.

Example 4

Find the equation of the circle with PI P2 as diameter where PI (- 1,2) and P2 (2, -3). Solution

Method 1 The centre of the circle is (T, v) = ( h , -a). The radius is '

1 1 -,/(-I - 2)2 + (2 - (-3))2 = -6. 2 2

If P(x, y) is any point on the circle then

which simplifies to


as the required equation of the circle.

Method 2 Let P(x , y) be any point on the circle. Then,

Y - 2 slope of PP2 = -. y + 3 slope of PP2 = - x + 1 ' 2 - 2 '

For perpendicularity of PPl and PP2.

SO ( y - 2 ) ( ~ + 3) + ( x + 1 ) ( x - 2) = 0 which simplifies to yield

~ ~ + ~ ~ - x + y - 8 = 0

as the required equation of the circle.

Exercises 2.2

Obtain the equation of the circle through the three points A(1,3), B(2, - 1 ) and C ( - 1 , l ) .

Given the circle x2 + y2 - 22 + 4y = 0 find the equation of its diameter which passes through the origin.

Find the point which is diametrically opposite t o ( 2 , l ) on the circle xa + y2 - 32 + 5y - 4 = 0.

Find the equation of the circle which passes through the points A(1,4) and B(3,7) and has the centre on the x-axis. [Hint: The centre of the required circle lies on the perpendicular bisector of AB.] Find the equation of the circle which passes through the origin and the point A(1, -5) and whose centre lies on the straight line 32 - y = 11 Show that the locus of the midpoint of the line joining the origin to the circle x2 + ya + 4x + 4y+4c = 0 , c < 2, is also a circle and determine the centre and the radius. [Hint: If M(h, k) is a typical midpoint then P(2h, 2k) lies on the given circle. See Figure 2.1.1

FIG. 2.1 Find the equation of the circle which passes through the origin and cuts off intercepts of 3 and -2 on the x- and y-axis respectively. Find the equation of the circle with ( -1 ,3 ) and (5,7) as the ends of a diameter.

Obtain the equation of the circle through the points A(-1, - l ) , B(2,4) and C(-2 ,8) . [The following hint gives a third method: Find the equation of the perpendicular bisector of AB. Call it el = 0. Next, find the equation of the perpendicular bisector of BC (or C A ) . Call it 4 2 = 0. Solve el = 0 and 4 = 0 simultaneously to obtain the centre of the desired circle. If the coordinates of this centre are K(h , k), compute the radius of the circle by computing IKAI, (KBI or IKCI. The required equation of the circle can now be written down.]

2.3 Intersection of circle ai.18 line 51

10 Obtain the equation of the circle with PlP2 as diameter where PI = (21, yl) and P 2 = (22, y2).

2.3 Intersection of circle and line

Let us consider the problem of determining the points of intersection of a circle and a line which passes through a given point (xo, yo). Except for the case of lines parallel to the y-axis, the line through the point (so, yo) is of the form

We shall treat the case of lines parallel to the y-axis separately. To find the points of intersection of the line through (zo, yo) given by equation (4) and the circle

we solve the two equations (4) and (5) simultaneously. We have chosen the circle with centre (0,O) for convenience.

Substitution of m(x - zo) + yo for y in (5) yields z2 + [m(z - XO) + yoI2 = r2 or

This is a quadratic equation in x. It may have two real roots (corresponding to the points of intersection of the line and the circle); or no real root (corresponding to the case where the line does not intersect the circle); or two coincident roots (in this case the line is called a tangent to the circle). See Figure 2.2a,b,c.

a. Equation (6) has two r e d roots b. Equation (6) has no real mots

c. Equalion (6) has two coincident mots

FIG. 2.2


2.3.1 Tangents to a circle

In the case above when the line ( 4 ) is tangent to wie circle (5) the discriminant of the quadratic in (6) vanishes, that is,

[2m(y0 - mxo)I2 = 4(1 + m 2 ) [ ( y o - mzo)' - r 2 ]

Equation ( 7 ) is a quadratic equation in m for determining the slopes of the two possible tangents to the circle ( 5 ) passing through the point ( x o , yo) . If A denotes the discriminant of this quadratic equation, then

A = 4 x i y ; - 4 ( r 2 - x i ) ( r 2 - y; ) = 4 r 2 ( x i + - r2) .

For A > O or > r2 there are two distinct roots of ( 7 ) , namely [ - x o y o f r JGim/ (r2 - x i ) . Note that x i + y: > r 2 implies that the point ( x o , yo) is outside the circle. In this case there are two distinct tanvnta. The equations of these tangents are given by

Y = -XoYo f T J ~

r2 - y; ( x - xo) + yo.

For A = 0, x i + = r2 there are two equal roots and hence only one tangent. In this case the point ( 2 0 , yo) is on the circle and

The equation of the tangent is 2 0

y = - - ( x - x o ) + yo Yo

The expression X ~ X + y l y - r* is olten denoted by T indicating its relationship with tangent. For the more general form of the equation of the circle, namely

the corresponding expression is

For A < 0, x i + Yi < r2 there is no real root and hence no tangent. Here the point is inside the circle.

Observe that if in equation ( 7 ) x i = r2 , then (7) is no longer a quadratic in m but a linear equat'ion which has one root for m. In this case, one of the tangents cannot be written in the form ( 4 ) . The other tangent may be of the type we excluded at the beginning; namely, a line parallel to the y-axis. Let us determine if this is the case.

A line parallel to the y-axis and passing through the point ( x o , yo) has the equation

2.3 Intersection of circle and line 53

This line intersects the circle z2 + yZ = r Z at the points where z$ + y2 = r2. That is,

For the line to be a tangent the two points of intersection ( x o , d m ) and ( z o l -d-) must coincide. That is d q = - d G g or r2 - x i = 0 or r2 = 28. This is the condition obtained above; and the equation of the tangent is x = xo.

Note that for the special case when (zo,2(o) is on the circle, yo = O and the equation (7) is therefore valid here,

Example 5

Obtain the point of intersection of the straight line 32 - y + 5 = 0 and the circle z 2 + y2 = 25.

Solutba To obtain the point of intersection of the line and the circle, we solve the two equations simultaneously. Rom 32 - y + 5 = 0 we obtain y = 32 + 5. Substitute this in x2 + y2 = 25 to obtain

i.e., 102' + 302 = 0 or z(z + 3) = 0 so that x s 0 or -3. Using y = 3 x + 5 we obtain y = 5 or -4. Thus, the points of intersection are ( 0 , 5 ) and ( -3 , -4).

Example 6

Obtain the equations of the tangents through ( - 2 , l l ) to the circle x2 + y2 = 25.

Solutba The equation of a line through ( - 2 , l l ) with gradient m is given by y - 11 = m(z + 2 ) i.e., y = mz + 2m + 11. This intercepts the circle t2 + y2 = 25 where x2 + ( m x + 2 m + 11)2 = 25 or

This equation has two coincident roots for x if and only if

Thus m = - 8 or T . Substitution of these values in the equation y = m x + 2 m + 11 gives the equations of the tangents as

4x +- 3y - 25 = 0 and 242 - 7 y + 125 = 0.

Disisru melhod Let the eqqotiop of tbe reqvired Lsngeato be or + bp +c = 0. We will use the fact that the distance from the centre of a circle to a tangent is the radius. From the result (15) of Subsection 1.4.3 the distance from the centre (0,O) to the line ax + by + c = 0 is c / d m . Thus we have


Since (-2, l l ) lies on the tangent we have

Solvinq equations (i) and (ii) simultaneously we obtain

Thus

. . , .-,',

(ii)

Substituting into (ii) we have

c 2 5 a 4 - - -- c 125 a 24 f o r - = - and - - - - b - 3 h 3 b - 7 b 7 '

for - = --

Therefore, as before, the equations of the tangents are

42 + 3y - 25 = 0 and 242 - 7y + 125 = 0.

Example 7

Find the equation of the two tangents from (4, -2) to the circle x2 - 22 + y2 - 8 = 0.

Solution

The line y = m ( x - 4 ) - 2 (*I

passes through the point (4, -2). This line intersects the circle x2 - 22 + y2 - 8 = 0 where x2 - 2x + [m(x - 4) - 2j2 - 8 = 0 or

This equation has two coincident roots if and only if

that is m = &. Thus we have one tangent line of the form (*) namely

The other tangent is a line parallel to the y-axis through the given point; i.e.,

Distance method We first determine the center and radius of the circle. To do this we complete the squares in equation x2 - 2x + Y2 - 8 = 0 to get (z - + Y2 = 32. Thus the circle has centre (1,O) and radius 3. Let


the equation of the tangent be ax + by + c = 0. The distance of the centre to this line is the radius and hence

, l a ( ' ) + b ( o ) + '1 = 3 or ( a + c )2 = 9(.2 + b2) d m

Since ( 4 , -2) lies on the line 4 a - 2 b + c = 0 .

( 9

(ii)

Eliminating c from equations (i) and (ii) we obtain

Simplifying this and solving gives 12

b = 0 or b = - - 5 a .

Substituting into (ii) we have 44

c = - 4 a or c = - - 5 a .

Consequently ax + by + c = 0 becomes

as before.

Alternative method

We now develop an alternative method for calculating the equations of the pair of tangents from an external point to a circle. Let P ( x l , y l ) be the given point outside the circle x2 -t y2 = r 2 (Figure 2.3).

FIG. 2.3

We want to find the equations of the two tangents which can be drawn from P ( x 1 , y l ) to the circle x2 + Y 2 = r 2 . Draw any line through P cutting the circle at A and B. Let Q ( x , y) be any

5 6 Coordinate Geometry of the Circle

point on this line and suppose the circle divides the line PQ a t the point A in the ration k : 1; The coordinates of A are then

k x + X I k y + yl

The point A lies on the circle x 2 + y2 = r2. Therefore,

This quadratic in k has two roots corresponding to the points of intersection of the line and the circle. For the line to be a tangent to the circle, the two roots will be coincident. Hence,

If we set f (x, y) = x 2 + y2 - r2 then the equations of the two tangents that can be drawn from the point P(x1, yl) to the circle x 2 + y2 - r2 = 0 is given by

Example 8

Find the equations of the two tangents from (3, -2) to the circle x 2 + y2 - 9 = 0. Solution

The equations of the two tangents are given by (9); namely,

where ( x l , yl) = (3, -2) and f(x, y) = x 2 + y2 - 9 ; r2 = 9. NOW, f(x1, y1) = 3 2 + (-2)' - 9 = 4 , and xxl + yyl - r2 3 2 - 2 y - 9. Hence, the required equations ate given by

so that, 4 x 2 + 4Y2 - 36 = 9 x 2 - 12xy + 4 y 2 - 5 4 2 + 36y + 8 1

Factorizing gives ( 5 2 - 12y - 39)(x - 3) = 0,

and the required equations of the two tangents are:

x - 3 = 0 and 5 x - 12y - 3 9 = 0.


Length d t angents from an external point t o a circle

Consider the circle S whose equation is given by x2 + y2 + 2hx + 2ky + d = 0, with centre (-h, -k) and radius r given by r = d h 2 + k2 - d. Let P ( T , . yl) be any point outdde the circle, S (Figure 2.4).

FIG. L4

Let Q1 and Q2 denote the points of contact of the tangent drawn from P(xl , yl) to the circle, S. We are interested in finding the lengths IPQl], lPQal of the line segments P Q l and PQ2 . Now, by Pythagoras theorem,

Thus, the equare of the length of a tangent from a point ( a l , yl) to a circle is obtained by the eubetituiion of the coordinates of the point in the equatiop of the circle, provided that the coefficients of z2 and y3 have been made unity.

Example 9

Find the length of the tangent drawn from P(-3,O) to the circle 3x2 + 3y2 - 122 - 4y - 4 = 0.

Solution Let t denote the length of this tangent. Making the coefficients of x2 and y2 unity in the given equation, we obtain

This circle has centre C(2, i) and radius 9 (Figure 2.5).

FIG. 3.3


Therefore,

Therefore

Note that the use of the formula derived above on (i) givcs

so that t = fi as before.

The normal t o a circle at a point Pl(x;, yl) on the circle is thc line perpendicular to the tangent to the circle at P l (x l , yl).

Consider the circle S given by S x2 + y2 + 2hx + 2ky + d = 0 with the point P l (x l , yl) on S. Now the tangent to the circle at any point on the circle is perpendicular to the radial line from that point. See Figure 2.4. Therefore, the normal to the circle at any point is an extension of the radial line to the point. The centre of the circle S is (-h, -k) and the slope of the radial line from Pl(x1,yl) is

1 - - - Y l + k 21 - (-h) - XI + h '

Thus, the equation of the normal to S at P l (x l , yl) is given by

Multiplying out and collecting terms in x and y we have

Example 10

Find the equation cif the normal to the circle xa + y2 - 42 - 4y - 5 = 0 at the point (0,5).

Solution

We complete the squares of the given equation for the circle to obtain (x - 2)2 + (y - 2)2 = 13. The centre of the given circle is (2,2). The normal at (0,5) is the line passing through the two points (2,2) and (0,5). The slope of this line is given by = -:. Thus, the equation of the normal is

2.3 Intersection of' circle and line 59

Ali ter

We can first differentiate the given equation of the circle implicitly to find the 2. Thus,

At (0,5), 2 = & = which is the slope of the tangent to the circle at (0,5). Thus, the slope of thc normal at (0,5) = -: and the equation of this normal is given by

3 y - 5 = --(x -0) or 3 x + 2 y - 10 = 0,

2 as before.

Example 11

Find the equation of the normal to the circle x2 + y2 = 16 at the point (-4,O).

Solution

'I'll(-: normal passes through the point (-4,O) and also through the centre of the circle which is (0,O). 'I'lms, the equation of the normal is the equalion of the line passing through the two points (-4,O) and (0,O). The slope of this h e is = 0. Hence, the equation of llris normal is given by

Ali ter

Differentiating x2 + y2 = 16 implicitly, we obtain

Since the normal is perpendicular to the tangent, its slope is t. At (-4, O), 2 = = 0, and the equation of the normal at (-4,O) is then given by

y - 0 = 0(x + 4) or y = 0,

as before.

2.3.3 Chord of contact

Given a circle anu a point f outside the circle we have seen that there are two tangents that can be drawn from the point to the circle. The cl~ord of the circle joining the two points of contact is called the chord of contact of the tangents. See figure 2.6.

FIG. 2.6


Let.us derive the equation of the line containing the chord of contact of the tangents through the point P(x l , gl) to the circle xa + y2 = r2;

Let PB and P C be the two tangents from P to the circle, where the coordinates of B and C are (x2, y2) and (x3, m) respectively (see Figure 2.6). Now the equation of P C in Figure 2.6 (since PC ie a tangent to x2 + y2 - ra = 0 a t C(x3, y3)) is given by

and the equation of P B is given by

Since (21, yl) lies on P B and atso lies on PC, equations (i) and (ii) become (respectively)

(ii)

(iii)

(i.1

R o m (iii) and (iv) we see that (xa, ya) and (xs, ys) lie on the line

Hence, the equation of the line containing the chord of contact B C of the two tangents from P ( x l , yl) to the circle x2 + ya = rz is

x x l + yyl = r2.

It is not difficult to deduce that with respect to the equation x2 + y2 + 2hx + 2ky + d = 0, thb chord of contact with respect to the point Pl(xl , yl) is given by

Given a circle and a point P the polar of the point with respect to the circle is the locus of points of intersection of the tangents at the ends of chords of the circle passing through the point. This point ie called the pole of the locus.

To determine the equation of the polar of the point P(x l , yl) with respect to the circle x2+y2 = r2, let (x*, y*) be any point on it. See Figure 2.7 for the three cases where the point P is (i) outside the circle, (ii) on the circle and (iii) inside the circle. The equation of the line containing the chord of contact is

2.3 . Intersection lof.circle ad line , 61

FIG. 2.7

By definition the line passes through P. Therefore

This shows that (x*, y*) and thus any point on the polar satisfies the equation

It is interesting to note that the equation xlx + yly = r2 represents (i) the equation of the tangent to the circle x2 + y2 = r2 at a point (XI, yl) on the circle, (ii) the equation of the chord of contact of tangent to the circle x2 + y2 = r2 from a point P(x!, yl)

outside the circle and (iii) the equation of the polar of P(xl , yl) with respect ot the circle za + y2 = r2.

Observe that cases (i) and (ii) are special cases of (iii).

Example 12

Let the straight line 5x - 39 + 1 = 0 intersect the circle x2 + y2 - x + v -- 2 = 0 at the points A and B. Obtain the coordinates of the point of intersection of the tangente to the circle at A and B.

62 ' Coordinate Geometry of the Circle

Solution

(A sketch is given in Figure 2.8).

FIG. 2.8

The centre of the given circle is (I, -$) and the radius is 4. Let Pl(x1,yl) be the desired point of intersection of the two tangents. Now, the equation of the chord of contact of the two tangents from P l ( x l , yl) to the given circle is given by

This represents the same straight line as 52 - 3 y + 1 = 0. Hence, their corresponding coefficients must be proportional. Therefore

1 1 1 x 1 - 5 y 1 + 5 - - p 1 + 4 y 1 - 2 -- - 5 - 3 1

Solving the simultaneous equations involved yields ( X I , yl) = (-2, l ) .

Example 13

If the chord of contact of the pair of tangents from a point P to the circle x 2 + y 2 = a 2 always touches the circle x 2 + y2 - 2 a x = 0, show that the locus of P is the curve,given by y2 = a(a - 22). Solution

A sketch is given in Figure 2.9.

FIG. 2.9

2.3 ' Intersectiorl of tircle and line 63

Let P ( x l , y l ) . The circle x 2 + y2 - 2ax = 0 has centre ( a , 0 ) and radius a. Now, equation of the chord of contact from P to x 2 + y2 = a2 is given by

2 x x l + yyl = a . (9

Since this chord is a tangent to the circle

x 2 + y2 - 2ax = 0, (ii)

then the distance of the centre of (ii) to the line (i) must be equal to the radius of (ii). Thus,

so that

or x ; - 2axl + a2 = x: + y:. Thus y? = a(a - 2x1). But ( x l , y l ) is a variable point. So, the locus of P is given by y2 = a(a - 2 x ) , as desired.

Exercises 2.3

Find the coordinates of the centroid of the triangle formed by 3x2+2xy-y2 = 0 and 3x+y-2 = 0 . [Hint The straight line pair passes through the origin.]

Obtain the equations of the tangents to the circle x 2 + y2 = 10 which are parallel to the line y = 3 x + 7 .

Obtain the equations of the tangents through the point ( 1 , 3 ) to the circle x 2 + y2 = 5.

If y = m x + 5 is a tangent to the circle x 2 + y2 = 5 , obtain the values of m.

Find the condition that y = me + c will intersect the circle x 2 + y2 = r 2 at two distinct real points.

Let the line y = m x + c cut the circle x 2 + y2 = r 2 at two points P and Q. Show that the

length of the chord PQ is 2 1 + m 2 '

Obtain the length of the tangents from the origin to the circle x 2 + y2 + 2x - 4 y + 4 = 0 .

Calculate the length of the tangents from (5 ,12 ) to the circle 2x2 + 2y2 = 69.

Find the equations of the tangents from ( 2 , -3) to the circle x 2 + y2 + 6z - 4 y + 8 = 0 .

Show that the pair of tangents from ( - 1 , 3 ) to the circle x2+y2 = 5 are mutually perpendicular.

Find the equations of the two straight lines which pass through the intersection of x - y + 2 = 0 , 2x - y + 3 = 0 and are at a distance of from the point ( 1 , l ) .

Find the equation of the tangent to the circle x 2 + y2 = 7 at the point ( 0 , a) on the circle.

Find the equation of the tangent t o the circle x 2 + y2 - 4 x - 6 y + 3 = 0 at the point ( 1 , 6 ) on the circle.


14 Obtain the chord of contact of the tangents to the circle x2 + y2 = 5 from the point (-5, -5) 'and hence determine the equations of the tangents from (-5, -5) t o the circle.

15 Find the chord ,of contact of the tangents to the circle x2 + y2 - 42 - 6 y + 3 = 0 from the origin and hence prove that the equation of the tangents is x2 + 12xy + 6y2 = 0 .

16 Prove that the chords of contact of the tangents to the circle x2 + y2 + 2hx + 2ky + d = 0 from the origin and ( h , k ) are parallel.

17 Obtain the coordinates of the points of contact of the tangents from (2 ,O) to the circle x2 + y 2 - 2 ~ + 6 y + 5 = 0 .

2.4 Systems of Circles

2.4.1 Orthogonal circles

Let S1 = 0 and S2 = 0 be two circles and suppose the circles intersect a t A and B (see Figure 2.10).

FIG. 2.10

Let el = 0 represent the equation of the tangent to the circle S1 = 0 at A, and l2 = 0 represent the equation of the tangent t o the circle S2 = 0 atA. The angle 0 between the two circles (Figure 2.10) is defined as the angle between the two tangents. F'rom geometry we can prove that the angles a t the two intersections are equal. If the two tangents el = 0 and l2 = 0 are perpendicular, the circles are called orthogonal. Observe that in this case the tangent to one circle passes through the centre of the other. To see this, observe that if 0 = 90° (Figure 2.11) then l l = 0 is perpendicular to l2 = 0 . ' Since el = 0 represents the tangent to S1 = 0 at A, any line perpendicular t o el = 0 a t A must pass through the centre of S1 = 0 . In particular, if the circles are orthogonal, t2 = 0 must pass through the centre of S1 = 0 . Similarly, l l = 0 must pass through the centre of S2 = 0 .

FIG. 2.11

Consider the two circles given by

~ 1 = x ~ + y ~ + 2 h z + 2 k y + d = 0 and ~ ~ + ~ ~ + 2 h ' x + 2 k ' ~ + d = 0 .

We want to find a condition for the two circles to be orthogonal. Let the two circles intersect a t P and Q. If the circles are orthogonal then the tangent to one circle passes through the centre of the other.

The centre of S1 = 0 is A ( - h , - k ) and the radius is given by rl = d h 2 + k 2 - d. Similarly, the centre of S2 is B( -h ' , - k l ) and its radius is given by 1-2 = dh12 + kI2 - d'. If the circles intersect orthogonally, then triangle P A B is right-angled at P , and so 1ABI2 = I P B ~ ~ + IpAI2 which yields

( h - h')2 + ( k - k1)2 = r: + r; or ( h - h1)2 + ( k - L ' ) ~ = ( h 2 + k 2 - d ) + hI2 + kI2 - d'

so that h 2 - 2hh1+ hI2 + k 2 - 2kk1+ kI2 = h 2 + k 2 - d + h I2+kr2 - d'

or

(11 ) [ 2hh1 + 2kk1 = d + d' I This is the required condition for the orthogonality of the two circles S1 = 0 and S2 = 0.

Remark 2.1

Suppose that the condition 2hh' + 2kk1 = d + d'

is satisfied. Add h 2 + hI2 + k 2 + kI2 to both sides of this equation to obtain

h 2 + hI2 + k 2 + kI2 + 2hh1 + 2kk1 = h 2 + hI2 + k 2 + kI2 + d + dl

which may be rearranged to give

( h - h')2 + ( k - k1)2 = ( h 2 + k 2 - d ) + hI2 + kI2 - d'

We have thus shown that

(i) if the circles are orthogonal then condition (11 ) must be satisfied, and

(ii) if condition (11 ) is satisfied then the circles must be orthogonal.

Condition (11 ) is therefore necessary and sufficient for orthogonality of S1 = 0 and S2 = 0.

Example 14

A circle passes through K ( a , b) and cuts orthogonally the circle x 2 + y2 + 2hx + 2 k y + d = 0. Prove that the centre lies on the line 2(a + h ) x + 2(b + k ) y = a2 + b2 - d.

Solution


FIG. 2.12


Let the equation of the required circle be given by

with centre ( - h i , - k t ) . This circle passes through K ( a , b) and so the coordinates of Ii must satisfy the equation of the circle. Thus, we have,

which yields 2ah' + 2bk' = -d' - a2 - b2.

If the two circles r u t orthogonally we have 2hh1 + 2kk1 = d + di from which we obtain dl = 2hh1 + 2kk' - d . Substitution of d' in (i) yields

( i i )

Since the centre of the required circle is ( - h i , - k t ) , it follows from (ii) that this centre satisfies the equation

2(a + h)x + 2(b + k ) y = a2 + b2 - d ,

as required.

Example 15

Find the condition for the chord of contact of tangents from a point P to the circle x2 + y2 = r 2 to subtend a right angle a t the origin. Also find the locus of the points of intersection of perpendicular tangents to the circle x2 + y2 = r 2 .

Solution

A sketch will be helpful and is given in Figure 2.13.

X

FIG. 2.13

The chord of contact is given by the line segment RQ, where the tangents from P(x1, yl) to the circle meet the circle a t R and Q . Observe that angle O R P = 90° and angle O Q P = 90' so that

2.4 Systems of Circles 67

lORl = lQPl = r and lOQl = lRPl = r and thus ORPQ is a square. Thus, the required condition is that the length of the tangent from P ( x l , yl) to the circle x2 + y2 = r2 is equal to the radius of the circle, i.e., lPQI = r or I P Q ~ ~ = r2 or x: + Y: - r2 = r2 which reduces to

as the required condition. Since the coordinates ( x l , y l ) of the point of intersection of perpendicular tangents to the circle

x2 + y2 = r2 satisfy the equation z2 + y2 = 2r2 and this equation is independent of the points on t,he circle at which the tangents are drawn, it follows that the locus of the points of intersection of perpendicular tangents to the circle x2 + y2 = r2 is given by x2 + y2 = 2r2.

Example 16

Prove that the midpoints of those chords of the circle x2 + y2+2hx+2ky+d = 0 which pass through a point P(x1, yl) lie on the circle ( x - x l ) ( x + h ) + (y - yl ) (y + k ) = 0.

Solution


FIG. 2.14 4.

Let a typical midpoint of one of the chords passing through Pl (x l , yl) be denoted by M(Z, a ) . Let I<(-h, -k) denote the centre of the given circle. Then K1M is perpendicular to MPl for any M .

Y - Y 1 ' + and Gradientof MPl = - Gradientof K M = - x + h 5 - 1 1

Since K M is perpendicular to MPl we have,

which implies that the midpoints lie on the circle

( X + h) (x - 2 1 ) + ( Y + k ) (y - Y I ) = 0,

as required.

0

68 Codrdinate Geometry of the Circle

2.4.2 Radical axis of two circles

Let P (x l , yl) be a fixed point in the plane. Suppose that a line through P B as shown in Figure 2.15.

FIG. 2.15

The product of the directed distances PA, P B is always a constant

meets a circle a t A and

and is called the power of the point, P , with respect to the circle. To verify this we shall use the parametric form of the equation of the line PAB. Let the circle be represented by

The parametric equations of the straight line P A B are given by

where 0 is the inclination of the line to the positive direction of the x-axis. Substituting these equations in the equation of the circle we obtain,

(xl + t cos e)2 + (yl + t sin e)2 + 2h(x1 + t cw e) + 2k(yl + t sine) + d = o or

t2 + 2t[(xl + h) cog0 + (yl + k)sinO] + X: + y: + 2hxl + +2kyl + d = 0.

This is a quadratic equation in t and if t1 and tz are the roots of this equation, then t l and t2 correspond to the distances JPAI and IPBI so that ltlt21 = IPAl. IPBI. F'rom the theory of roots of quadratic equations we obtain immediately that

Note that lPAl . lPBl is a constant and for the special case of the tangent (A and B coincide) we recover the square of the length of the tangent derived earlier.

Remark 2.2

The power of P (i.e., lPAl . IPB[) is positive or negative according as P is outside or inside the circle. See Figure 2.16 for P inside the circle, where P A and P B are necessarily drawn in opposite directions. If P lies on the circle then 1 PA1 . 1 PBl = 0, and in this case the power of P is zero.

FIG. 2.16

2.4 Systems of Circles 69

Consider the two circles given by S1 = 0 and S2 = 0, and let P ( x l , y l ) be a point in the plane.

FIG. 2.17

Suppose a straight line from P intersects S1 = 0 at A and B , and another straight line from P intersects Sz = 0 at Q and R as in Figure 2.17. It may turn out that the powers of P with respect to the two circles are equal.

Suppose P(x1, yl) is any point whose powers with respect to the two circles

are equal. Let a straight line from P intersect S1 = 0 at A and B and, another straight line from P intersect S2 = 0 at Q and R. Since the powers of P with respect to S1 = 0 and Sz = 0 are equal, we must have lPA19 (PBI = ( P Q l . IPR(. But

and JPQJ . lPRl= x: + y: + 2hrx1 + 2kry1 + dl

so that )PA/ . J P B l = IPQl . [PRl implies

which reduces to 2(h - hl)xl + 2(k - kl )y l + d - dl = 0.

Thus, P ( x l , y l ) lies on the straight line given by

We may write this equation simply as S, - S2 = 0.

If we can locate all points in the plane such that their powers with respect to the two circles are equal, the locus of all such points is called the radical axis'of the two circles. More formally, we have the following definition.

DEFINITION The straight line which is the locus of all points P in the plane such that their powers with respect to two circles are equal is called the radical axis of the two circles.


Example 17

Let S1 = 0 and Sz = 0 be two circles which intersect at A and B and let P be a point on the line B A produced (see Figure 2.18).

FIG. 2.18

Then it is clear tha t lPAl lPBl is the same for both circles and so the power of P with respect to the two circles is the same. So, P lies on the radical axis of the two circles. I t is easy to see that the radical axis of the two circles is, in fact, the straight line B A produced, i.e., the common chord of the two circles.

R,emark 2.3

If SI = 0 and Sz = 0 are two circles, the radical axis is always perpendicular to thc 1i;e joining the centres of the two circles. For, from equation ( I ) , gradient of the radical axis is -% and the

gradient of the line of centres is H. R.emark 2.4

If the two circles towh , i t then follows that their radical axis is the common tangent a t the point of contact.

T h e concept of radical axis will be useful in the next subsection.

2.4.3 Coaxial circles

If a system of circles is such that the radical axis of one pair is the same as that of any other pair, then the circles are said to form a coaxial system.

Example 18

All circles that pass through two fixed points A and B have the line segment A B as their common

chord, and so all such circles form a coaxial system. See Figure 2.19.

' 4 Radical axis

Line of centres

FIG. 2.19

Example 19

All circles that touch at one point form a coaxial system with their colnmon tangent as their radical axis. See Figure 2.20.

Line of centres

1 Radicalaxis

FIG. 2.20

We have remarked that the radical axis of two circles is perpendicular to the line joining the centres of the two circles. I t then follows that the centres of the circles of a coaxial system lie on a straight line.

Suppose now we take as z-axis the line of centres and as y-axis the common radical axis for a coaxial system. Then the equations of any two circles of the system may be taken as (since the y-coordinate of the centre of each circle is zero).

Then the radical axis is given by S1 - S2 = 0, i.e., 2(hl - h2)x + dl - d2 = 0. If the radical axis is the y-axis, then its x-component is zero. This implies, from the above equation, that dl - d2 = 0 or dl = dz. Thus, the equation

x 2 + y 2 + 2 ~ x + d = 0 ,

for varying X and constant d, represents a coaxial system of circles with centres on Ox, and with radical axis as Oy.

72 Cuordinate Geometry of the Circle

Exercises 2.4

Use the methods of this section to solve the following problems.

1

2

3

4

5

2.5

Let

Write down the equation of the circle through the origin whose centre is (a , b). Prove that the tangent at the origin is ax + by = 0. [Hint Use the fact that tangent is perpendicular to the radius at the point of contact.]

If ax + by = c touches x2 + y2 = r 2 , find its point of contact. ra [Hint The equation xxl + yyl = r2 and ax + by = c are the same line if 2 = 9 = -;.I

Show that the chord of the circle x2 + y2 = ra whose midpoint is M ( x l , yl) has equation 2x1 + yy1 = x? + y;. [Hint If N is any point of tlrc required chord and 0 is the centre of the given circle then O M is perpendicular to MN.]

Find theequation of the circle orthogonal to x2+y2+4x-4y-2 = 0 and x2+y2+2x-2y-1 = 0, and whose centre lies on the line 22 - 39 - 2 = 0.

Find the equation of the circle orthogonal to xa + y2 + 62 - 7 = 0 , x2 + y2 = 1 and x2 + y2 - 42 - 4y - 5 = 0.

Equations of the forms S + AL = O ; S + AS' = 0

For all values of A, consider the equation

S + A L = O ,

This is written compactly as

~ ~ + ~ ~ + 2 h x + 2 k ~ + d + X ( l x + r n y + n ) = 0

which represents a circle provided

sinct , (i) the coefficients of x2 and y2 are the same ,

(ii) there is no term in xy (iii) the equation is a quadratic in x and y.

The equation S + AL = 0 represents the equation of the circle passing through the intersection of S = 0 and L = 0.

Next let

2.5 Equations of the forms S + AL = 0; S + AS' = 0 73

For all values of A # -1, consider the equation S + AS' = 0. This is written explicitly as

or (1 + A)x2 + (1 + A)y2 + 2(h + Ah')x + 2(k + Akl)y + d + Ad' = 0,

which represents a circle for A # -1. The equation S + AS' = 0 represents the equation of any circle passing through the points of intersection of the circles S = 0 and S = 0, where A is to be determined. Observe that if A = -1 then S + AS' = 0 reduces to S - S' = 0 which is no longer the equation of a circle, but that of the radical axis of the two circles.

Several examples given below can be solved by some techniques studied earlier. We shall, however, use the methods established in this section to solve them.

Example 20

Find the equation of the circle passing through the points A(x1, yl), B(x2, y2) and C ( X ~ , y3). See Figure 2.21.

FIG. 2.21

Solution

First, find the equation of the line AB. Call it L = 0. Next, find the equation of the circle with AB as diameter. Call it S = 0. Then S+ AL = 0 represents the equation of a circle passing through AB since A and B are the points of intersection of L = 0 and S = 0. Finally, use the fact that C(x3, ga) must lie on the circle S + AL = 0 (and therefore must satisfy this equation) to find the value of A .

For a specific example, let us find the equation of a circle passing through the points A(2,6), B(4, -2) and C(-2,2). Now, slope of AB = = -4. Therefore, equation of A B is given by

Set L 42 + y - 14 = 0. Equation of the circle with AB as diameter is given by

Set S x2 + y2 - 62 - 4y - 4 = 0. Then, the required equation is of the form S + AL = 0 for some constant A. Now S + AL = 0 implies,

The point C(-2,2) lies on this circle and therefore the coordinates of C must satisfy this equation. so,

(-2)' + (2)' - 6(-2) - 4(2) - 4 + A(-8 + 2 - 14) = 0


or 8 - 20X = 0 so that X = g. Thus, the required equation is given by

Example 21

Find the equation of the circle with the chord of intersection of S r x2 + y2 + 42 - 2 y + 2 = 0 and the line L z x - y + 1 = 0 as diameter.

Solution

Any circle passing through the points of intersection of S = 0 and L = 0 has an equation of the form S + XL = 0, i.e.,

~ ~ + y ~ + 4 x - 2 ~ + 2 + X ( x - ~ + 1 ) = 0 , ( 9

where X has to be determined. See Figurt, 2 .22 .

L = 0

L, X

Required circle with AB as diameter

FIG. 2.22

The centre of the required circle lies on AB. But the centre of the required circle is (from (i)) given by

Since this lies on the line AB (i.e., on x - y + 1 = 0) we must have

We can now solve for X to obtain X = -2 and substitute in (i) to get

as the desired equation.

2.5 Equations of the forms S + XL = 0 ; S + AS' = 0 7 5

Example 22

The circle x2 + y2 + 2x - 4 y - 11 = 0 and 'the line x - y + 1 = 0 intersect at A and B. Find the equation of the circle through A, B orthogonal to the given circle.

Solut ion

Any circle through the intersection of the circle and the line is given by

For this circle to be orthogonal to the given circle we must have

or X = -16. So, the required equation is given by


Find the values of k which makes the straight line 3x + 4 y + k = 0 a diameter of the circle x2 + y2 + 8x + 6 y + c = 0 for all c .

Find the equation of the circle on the intersection of x2 + y2 = 25 and x + y = 4 as the ends of a diameter. Find the circumcircle of the triangle formed by the three straight lines 2 2 + y - 2 . = 0 , x - 3 y + l = O a n d x - 2 y - 2 = 0 . The straight line y = m x + c cuts off a chord of length 2X from the circle x2 + y2 = a 2 . Show that c2 = ( a 2 - X2)(1 + m2) .

Find the equations of the two circles of radius fi with their centres on the x-axis which touch the line x + y + 1 = 0 . Find the value of b if the equation 2x2 - 5 x y + 2 ~ ' - 7 2 + l l y + b = 0 represents two straight lines. For this value of b show that the two lines intersect at a point on the circle x2 + y2 - 122 + 6 y + 20 = 0 , and find the equation of the tangent to the circle a t this point. The circles x2 + y2 + y - 4 = 0 ; x2 + y2 - 31: - 5 y + 2 = 0 are two members of a coaxial system; the circle x2 + y2 - y - 2 = 0 is a member of another coaxial system, of which the radical is the line x + y - 1 = 0 . Show that the two systems have a common circle and find its equation.

The tangents drawn from a point P on the x-axis to the circle x2+ y2 - 2ax+2a- 1 = 0 , ( a > 1 ) touch the circle a t Q and R. Prove that the point P can be found such that QR subtends a right angle at the origin only if a > 2 + fi. Find the values of X and p for the equation Xy(3x-y)+py(3x+y-6)+(3x-y)(3x+y-6) = 0 t o represent a circle. Hence, find the circumcircle of the triangle OAB whose sides OA, OB, AB have the equations y = 0 , y = 3 x , 3x + y = 6 respectively. Obtain also the coordinates of the centre of the circle which cuts this circle orthogonally at 0 and A.

76 Coordinate Geometry of the Circle r l

. , j ' :.

10 Obtain the equation(s) of the pair of tangents which can be drawn from the origin t o the e i h x2 + y2 + 82 + 6y + 21 = 0, and calculate the angle between them.

11 P is a point which moves such that its distance from S(a,O) is k times its distance from S1(-a, 0). Show that the locus of P is the circle x2 + y2 - 2Xax + a2 = 0, where X = (1 + k2)/(1 - k2), k # 1. (This circle is called the circle of Appolonius).

12 Obtain the equation of the circumcircle of the triangle whose sides have the equations 22 + y = 2 , ~ - 3 y + l = O a n d ~ - 2 y = 2 .

13 A circle whose centre lies on the line 4y = x + 7 cuts the z-axis in points where x = 2 and x = 6. Find the equation of this circle. Show that the circle does not cut the y-axis in real points.

14 If the circle S xa + y2 + 6y - 16 = 0 and the straight line L = x + 2y - 4 = 0 intersect a t A and B, obtain the equation of the circle which is orthogonal to 5' at A and B.

15 Prove that the condition for the line lx + my + n = 0 t o be a tangent t o the circle (x - a)2 + (y - b)2 = r2 is that (la + mb + n)2 = r2(12 + m2). [Hint: The line is a tangent to the circle if the distance of the centre of the circle from the line is equal to the radius of the circle.]

16 Write down the equation of the polar of the point P ( a , b) with respect to the circle x2 + y2 + 2Xx + c = 0, and show that, if b # 0, then for every value of X the polar passes through a fixed point P*, find the coordinates of P*. Show that the circle on PP* as diameter is orthogonal to every circle of the given coaxial system.

17 Find the equation of the circle with the chord of intersection of S x2 + y2 + 2hx + 2ky + d = 0 and the line L l z + my + n = 0 as diameter.

18 The circle S G r 2 + y2 + 2hx + 2ky + d = 0 and the line L lx + my + n = 0 intersect a t A and B. Find the equation of the circle through A and B orthogonal to the given circle.

Conic Sections 3.1 Introduction

After the straight line and circle, the conic sections (popularly known as "conics" for short) are the next geometric curves in order of complexity. Their equations are of the second degree in two variables, namely

where the coefficients/parameters a , b, c, dl e, f are all real numbers with a t least one of a, b and c non-zero. The study of the conics was motivated primarily by the fact that they are the paths traced out by one planet moving near another (for example, earth round sun, moon round earth, Harley's comet) or even an electron moving round a nucleus.

In spite of this association with celestial mechanics, the conics acquired their generic name "conic sectionsJJ fro111 :I more mundane consideration. Take two right circular cones and place them tip to tip so that their :)yes are colliriear as in Figure 3.1.

FIG. 3.1 Two cones placed tip to I ip with axes collinear

Now take any plane that intersects t h cones. In general, the cross-sectional view of the cut shows a curve. Depending on the position of t lw intersecting plane, the curve will be a parabola, an ellipse, a hyperbola or a point. These various c.iluations are illustrated in Figure 3.2 (i) to (iv). In other positions the cross-sectional figure may be a circle, line or two intersecting lines. Because each conic can be generated by ;I hcction or cut thronch cones, they art' called "conic sections".

(i) Ellips. (ii) Point,

78 Conic Srctions

( 1 1 1 ) Paral)ola FIG. 3.2 Some coliic sections

3.2 The Conic as a Locus

We shall obtain the equation of each of the conic sections in a "standard" form beginning from a common definition of these conics as loci traced out by moving points.

3.2.1 Definition and Examples

In order to obtain an equation for a conic we need a characteristic geometric property of the conic which can then be expressed in algebraic terms.

DEFINITION 3.1 A conic section is the curve traced out by a point P which moves 271 a plane in such a way that the ratio of its distance from a fixed point F to its distance from a fixed line (not containing F ) is a constant.

The fixed point F is called its focus, the fixed line its directrix and the constant ratio its eccentricity. The eccentricity is usually denoted by the letter e.

Example 1

Find the equation of the conic section with focus at the point ( - 1 , 2 ) and eccentricity e = if its directrix is the line 3a: + 4 y = 13.

Solution

Let P(x, y ) be a point on the conic. \Ve use ( E , q ) initially to avoid any possible confusion with (x, 9) in 3x + 4y = 13, the equation of the directrix. Srr Figure 3.3.

FIG. 3.3

3.2 The Conic as a Locus 79

Distance of P from focus = lPFl = J ( [ + 1 ) 2 + (1) - 2)2

Distance of P from directric = ( P K ( = 134 + . 4 ~ - 131 - 13E + 4q - 131 - dFTT2 5

But J P F J = e lPKl , therefore (PFI2 = e2(PK(' ie.,

Thus 5(e2 + 2E + 1 + v2 - 471 + 4) = 2(9t2 + 16772 + 24tq - 78E - 1041) + 169)

Using x , y instead of E , q we have

Example 2

Find the equation of the conic with directrix y = 22 + 3, its focus at the point ( 1 , l ) and which passes through the point ( 3 , 2 ) .

Solution

We do not know the eccentricity of the conic. Let it be e. Suppose the point P ( x , y) lies on the conic, then

Since the conic passes through the point ( 3 , 2 ) , we have

'Therefore

Hence the conic has the equation

which upon simplifying gives

80 Conic Sections

Example 3

Find the equation of a conic which has the line y = 22 + 3 as directrix, has its focus on y = x, e = 5/7 and passes through the point (3,2).

Solution

The only unknown is the focus which lies on the line y = x. Thus its x and y coordinates are equal. Suppose, therefore, that the focus has the coordinates ( a , a ) . Then

2

(2 - a )2 + (y - 5

or (x -a)=+ (y- a )2 = -(2x - +3)? 49

Since it passes through the point (3,2) we have that

Simplifying gives a2 - 5 a + 4 = 0. Therefore ( a - l ) (a - 4) = 0. Hence a = 1 or 4.

a = 1 gives the conic 29x2 + 44y2 + 20xy - 1582 - 68y + 53 = 0.

of our Example 2.

which simplifies to 29x2 + 44y2 + 20xy - 4522 - 362y + 1523 = 0.

Exercises 3.2.1

Obtain the equation of the conic satisfying the following conditions.

1 Directrix 22 - y + 1 = 0, focus (O,O), eccentricity e = ;. 2 Directrix x + y + 4 = 0, focus (1, I), ecceiitricity e = 1.

3 Directrix 3x - 4y + 5 = 0, focus (0, -1)) eccentricity e = 2.

4 Directrix y = 4, focus (1,2), eccentricity e = $. 5 Directrix y = 1, e = 1, focus lies on the line x = 2 while the conic passes through the point

( 5 , O . - 6 e = 3/5, the directrix is y = 9; focus lies on x = 4 and conic passes through the point (4,8).

7 Directrix is the y-axis, focus at (c, 0) and eccentricity is e .

3.2.2 Standard Forms for Conics

It turns out that to sketch the above conics and any other conic, it is generally more convenient to write them in a form in which there is no cross product, xy, term. This is achieved by using a special set of axes-those parallel and perpendicular to the directrix.

3.2 The Conic aa a Eocus 01

Let the directrix be the y-axis and let the x-axis pass through the focus (see Figure 3.4) . Suppose the focus is at the point ( c , 0). The point P ( x , y) lies on the conic with y-axis as directrix, F as focus and eccentricity e if

Thus

FIG. 3.4

To put equation ( 2 ) into standard form we need to complete squares.

Case I: 1 - e2 # 0

Divide equation ( 2 ) by 1 - e2 to obtain

so that

a. If 1 - e2 > 0 i.e. e2 < 1 or 0 < e < 1 ( e > 0 by definition), putting

82 Conic Sections

so that

, equation (3) becomes

Fkom (4a) the focus .is now the point

One further simplification can be made. Let X = x - a le and Y = y. Then equation (6a) becomes

where b2=a2(1 -e2 ) and O < e < l .

This is the standard equation of an ellipse.

b. If 1 - e2 < 0 i.e. e2 > 1, equation (3) becomes

with

so that

By (9), the focus is at the point

Once again with new (X, Y) axes where x + a/e = X and y = Y we obtain the form

where b2 = a2(e2 - 1) and e > 1.

3.2 The Conic as a Locus 83

This is the standard equation of a hyperbola. Case II: If 1 - e2 = 0, equation (2) reduces to

Putting $ c = a yields

So that the focus is at the point

(13b) ( c , 0) = (2a, 0).

in the new axes (XI Y) where X = x - a, Y = y, equation (12) becomes

where e = 1. This is the standard equation of a parabola.

DEFINITION 3.2 F o r the s tandard equations of a n ellipse a n d a hyperbola, t h e point X = 0,Y = 0 i s called t h e centre o f t h e conic. It i s called t h e ver t ex in t h e parabola.

Example 4

By putting the following conics in standard form determine the type of each conic. Find the coordinates of the centre or. vertex of the conic as applicable.

(i) 4x2 + 3y2 - 162 + 6y + 7 = 0,

(ii) 12x2 - By2 + 12x + 24y = 21, (iii) 4y2 - 32 - 8y + 10 = 0.

Solution

(i) 4x2 + 3y2 - 162 + Gy + 7 = 0.

Rearrange and complete squares of the x and y parts;

4(x2 - 42) + 3(y2 + 2y) = -7

4(x - 2)' + 3(y + = -7 + 4(-2)' + 3(1)' = 12.

Therefore 4(x - 2)2 3(y + 1)2

12 + 12 = 1

This is equivalent to

with X = x - 2 and Y = y + 1. This is therefore an ellipse with centre at the point X = 0, Y = 0; that is, x -2 = 0 and y + 1 = 0 or at (2,-1).

84 Conic Sections

(ii) 12x2 - By2 + 122 + 24y = 21.

12(x2 + x) - 8(y2 - 3y) = 21. Complete squares we have

Therefore,

1 2 ( ~ + $ ) ~ - 8 ( y - t ) 2 (x + f ) 2 - (y - ;)2 = 1 or 6 6 1 = 1 - -

2 4

which is equivalent to

with X = x + and Y = y - ;. Therefore, the equation represents a hyperbola with centre (-+, i). (iii) 4y2 - 32 - 8y + 10 = 0.

4y2 - 8y = 32 - 10 since there is no term in x2. i.e. 4(y2 - 2y) = 3x - 10. Therefore by completing squares we have

which yields 3

(y - 1)2 = -(x - 2). 4

This is in the standard form 3 y 2 = -X 4

with X = x - 2 and Y = y - 1. I t therefore represents a parabola with vertex a t ( 2 , l ) .

Remark 3.1

Note that in completing squares and in defining X and Y the coefficients of x and y must each be made unity. Thus we always have X = x - a, Y = y - /3 for some real numbers a, /3.

Exercises 3.2.2

P u t the following conics into standard form and determine the type of each conic. Find also the coordinates of the centre or vertex as appropriate.

1 12x2 + 8y2 + 12x - 24y + 15 = 0 2 4x2 - 3Y2 + 6x - 16y - 19 = 0

9 Show that the line y = m x + c intercepts the ellipse $ + $ = 1 only if c2 5 b2 + a2rn2 for all m. If (5,jj) are the coordinates of the midpoint of the chord of the ellipse, show that m a z y + b25 = 0.

3.3 Geometry and Sketches of the Conics 85

10 Show that the midpoint ( 2 , y) of focal chords of the parabola y2 = 4ax lie on another parabola. What is the equation of t h ~ s second parabola?

11 How would you use the definition of a parabola to construct one?

12 Use parametric methods to prove (i) I PFi I + 1 PFll = 20 for the ellipse $ + $ = 1. and (ii) IIPFi I - IPFzjj = 2a for the hyperbola - $ = 1.

3.3 Geometry and Sketches of the Conics

In this section we shall learn how to sketch each of the conic sections using its equation in the standard form given by equation (7), (11) or (14). We shall also identify the positions of the directrix and focus of each conic in terms of the new variables (X, Y) of these equations. First, however, we shall derive the meaning of these new variables.

3.3.1 Translation of Axes

Think of a coordinate system Oxy as two sticks which cross at right angles at the origin 0 at which point the sticks are rigidly fixed together. S~~ppose now we move this rigid frame about in the plane.

\

a. Translation alone . l totat io~~alone

Figure 3.5 illustrates the types of motion pos.;ible.

A Y'A

t 0' + X

0 c. Rotation aird trailslation

FIG. 3.5 Possible motions of a rigid coordinate frame

0

A new position of the rigid frame is shown as O'x'y'. In Figure 3.5a the origin is moved but the arms of the frame do not turn or rotate. This is called translation and the two axes Oxy, O'x'y.' are said to be parallel. When the origin is not moved so that only the arms turn, we have a rotation as

0' X

86 Conic Sections

in Figure 3.5b. The most general type of motion is shown in Figure 3 . 5 ~ and is a combination of rotation and translation.

Let Oxy be a coordinate system (see Figure 3.6) and the points 0', P have coordinates ( a , @ , (x, y) respectively.

Zone3

Zone 7

Zone 4

FIG. 3.6

Zone2

We shall show that the new variables (X ,Y) where X = x - a and Y = y - P are the coordinates of the point P relative t o a set of parallel coordinate axes OIXY with origin at 0'.

Consider then the two sets of parallel coordinate axes Oxy, O'XY with origins at 0,01 respectively. The Ox, O'X axes are parallel and so are the Oy, OIY axes. These axes divide the plane into nine regions or zones as shown in Figure 3.6.

P(x,Y>

Zone 1

0' Zone 5

X - a x i s

x - axis

, X (a$) zone (j

0 ' x * B

FIG. 3.7 'I'ranslation of axes

Suppose P is any point in Zone 1 of the plane. Let the coordii~at~es of P relative to Ozy be (x, y) and relative to OIXY be (X , Y). Let the perpendicular from P to the z-axis cut the O'X-axis a t A and the Ox-axis at B. See Figure 3.7. By definition O'A = X and O B = x. If the Y-axis cuts the x-axis at C , then O C = a. From Figure 3.7, O B = O C + CB. Thus x = a + X or X = z - a. In a similar way we can see that Y = y - p. It may be verified that when P lies in any of the ot,hcr zones, X and Y are still given in magnitude and direction by these formulae. Thus the coordinates (2, y), (X , Y) of any point relative to parallel axes are given by

where ( a , p ) are the coordinates of the origin of the XY-axes relative to that of the xy-axes.


We see therefore that the standard equations (7), (11) and (14) of the conics are in terms of parallel axes through their centres (ellipse and hyperbola) or vertex (parabola).

Example 5

Find the equation of the curve

(i) 2x + 3y = 7 (ii) 2x2 - 3Y2 + x - y + 4 = 0

in terms of the variables ( X , Y) of a parallel coordinate system through the point (-1,2).

Solution

For parallel axes through (cr,P), X = x - Q,Y.= y - ,@. Here (a, ,@) = (-1,2). Therefore, X = x + l , Y = y - 2 s o t h a t x = X X 1 , y = Y + 2 .

(i) Replace x , y in the equation using 3: = X - 1, y = Y + 2. Therefore 2x + 3y = 7 becomes 2(X - 1) + 3(Y + 2) = 7. This reduces to 2X + 3Y = 7 + 2 - 6 = 3.

(ii) By the same substitution, 2x2 - 3y2 + x - y + 4 = 0 becomes 2(X - 1)2 - 3(Y + 2)2 + (X - 1) - (Y + 2) + 4 = 0 which after simplification reduces to 2X2 - 3Y2 - 3X - 13Y - 9 = 0.

3.3.2 The Parabola

By (12) to (14), the equation of the parabola in stanrlianl T ; l r r ) ~ ii

(i) The parabola passes through the origin (0,O) of the X-Y plane. (ii) Y2 >_ 0 j a X 2 0. Therefore X 2 0 for a > 0 and X 5 0 for a < 0. Thus the parabola lies

only on one side of the Y-axis, depending on the sign of a . (iii) As a x + +co,Y2 -+ co, thus Y + f c o . The parabola goes off to plus or minus infinity and

is open in one direction. See Figure 3.8. (iv) If Y -is replaced by -Y in the equation of the parabola, (*), that equation remains unchanged.

This shows that for each X , both (X , Y) and (X, -Y) lie on the curve. Hence the parabola is symmetric about the X-axis. Where are the focus and the directrix of the parabola in terms of our standard variables (X, Y)? First, the focus.

Y A I I

I I I I I I

I I

X

FIG. 3.8 Sketch of the parabola in standard form I IDirectdx

(v) From (13b) the.focus of the parabola is a t (2a1 0) in the z-y coordinates. I11 the new XY-axes this is Y = y = 0, X = x - a = 2a - a = a . Thus the focus is at the point ( a , 0) in the X-Y coordinate system as located in Figure 3.8 (i) and (ii).

88 Conic Sections

(vi) To locate the directrix we note that the directrix is the y-axis whose equation is x = 0. In the new coordinate system X = r -a and the equation x = 0 becomes X = 0 - 0 = -0. Thus the directrix now has the equation X = -a. R'ote that the focus and the directrix are on opposite sides of the origin (See Figure 3.8).

Example 6

Sketch the curves (i) Y2 = % X (ii) X 2 = 2Y. In each case find the coordinates of the focus and the equation of the directrix.

Solut ion

(i) 4a = i. Therefore a = & > 0. This is like figure 3.8(i). The directrix has the equation X = -a = -& and the focus is at ( ago ) = ($,0) .

(ii) X 2 - 2Y + X 2 = 4aY. Thus 4a = 2, so that a = 4 > 0. However, the roles of X- and Y-axis are interchanged. Y > 0 and X -+ f oo (there is symmetry about the Y-axis) The parabola now opens upwards since y > 0. The directrix is given by Y = -a = -h . See Figure 3.9.

FIG. 3.9 Thc ~~ ; l r ;~bo la X 2 = 2Y

3.3.3 The Ellipse

By equation (4) to (7) the equation of the ellipse in standard form is

where b2 = a2(1 - e2) and 0 < e < 1. Its centre is at the point X = 0,Y = 0, the origin of the X-Y coordinates.

(i) Y = 0 =+ X 2 = a2 i.e., X = f a , intercepts on X-axis. X = 0 =+ Y2 = b 2 , Y = f b, intercepts on Y-axis.

.. X 2 Y2 Y (11) ; ; i -=l-- b.2 > O + - < l i . e . , - IYlSb.

b2 - Similarly IX I 5 a. Hence the ellipse lies entirely within the rectangle -a 5 X 5 a, -b 5 Y 5 b and is a closed curve.

(iii) Since replacing X by -X in the equation of the ellipse in standard form does not alter that equation then for each Y both (-X,Y) and (X,Y) lie on the ellipse. Hence the ellipse is symmetric about the Y-axis. Similarly it is also symmetric about the X-axis.

(iv) As 0 < e < li $ = 1 - e2 < 1. Therefore b < a. thus the length of the intercept on the X-axis is greater than that on the Y-axis. The X-axis is called the major axis and the Y-axis is called the minor axis of the ellipse. The ellipse with X-axis as major axis is sketched in Figure 3.10a.

3.3 Geometry md Sketches of the Conics 89

By convention the letter a in the standard formula always represents the length of the intercept on the major axis. Where the ellipse intercepts the major axis is called a vertex of the ellipse.

FIG. 3.10 Sketch of ellipse in standard form

YA

In our Example 4(i) of Subsection 3.2.2 we have

Y A

Since 4 > 3, the Y-axis ia the major axis (with an intercept of length 2) and thus we write the eauation as

a

Such a conic is illustrated in Figure 3.10b. Its vertices are a t (0, f a ) . (v) We note that the directrix is the y-axis whose equation is x = 0. In the new coordinate system

X = x - f and the equation of the directrix becomes X = -%. By the symmetry of the ellipee about the Y-axis, there are two directrices each a t the same distance f from the centre and perpendicular to the major axis of the ellipse (Fig. 3.11). Their equations are therefore X = & f . Note that % > a since e < 1.

(vi) To locate the focus of the ellipse we note that by equation 3.6b the coordinates of the focus are a = 41-ea) g = (q, 0). In the new X, Y-cordinates these become Y = y = 0, X = x - 7 e e

-ae. Again by symmetry about the Y-axis there are two foci a t ( f ae, 0). Since 0 < e < 1, ae < a and the foci are inside the ellipse and on the major axis (See Fig. 3.1 1).

I

b

-a a * x

* X b

-b

-a a. rriajor axis is X-axis b. major axis is Y-axis

:\ Directrix FIG. 3.11 Sketch of ellipse with foci and directrices

90 Conic Sections

Example 7

Sketch the ellipse

X 2 Y 2 (i) - + - = I

25 16 , P Y 2

(ii) - + - = 1 3 4

Find the coordinates of the foci and vertices and t h e equations of the directrices.

Solution

(i) Here a2 = 25, b' = 16. Therefore b' = a2(1 - e2 ) implies tha t e2 = 1 - 5 = 1 - = &. Thus e = g. T h e X-axis is the major axis. The foci are therefore a t ( r tae , 0) = (f 3 . g , 0 ) = (f 3,O). T h e vertices are (f a,O) = ( k 5 , O ) . T h e equations of the directrices are X = k:, i.e., X = k ? . T h e sketch is as in Figure 3.10a or 3.11.

X 2 Y 2 (ii) - + - = 1.

3 4 H e r e a 2 = 4,b2 = 3 s o that e2 = 1 - 5 = 1 - $ = i. T h u s e = a. a e = 2 . 3 = 1 ' e = 2 / 4 = 4 . Therefore since the major axis is the Y-axis, the foci are on it a t the points ( 0 , k a e ) = ( 0 , k l ) . T h e vertices have coordinates ( 0 , & a ) = ( 0 , k 2 ) . Also the equations of the directrices are Y = k t = k 4 . T h e sketch is as in Fig. 3.10b.

Finally, we derive a result which gives a second geometric characterization of the ellipse.

Example 8

x2 Y 2 For any point P on the ellipse - + - = 1 , a > b, show that the sum of its distances froin the two

a2 b2 foci Fl and F2 is equal t o 2a, the length of the major axis; tha t is,

Solution x 2 y2

Consider the ellipse - + - = 1. Let Fl(-ae,O), F2 (ae ,0 ) be its foci and P ( x , y ) any point on the a2 b?

ellipse.

FIG. 3.12

Draw the line ICPL through P and perpendicular to the directrices x = f % (see Figure 3.12). T h e coordinates of K, L are ( - 2 , y ) and ( f , y ) respectively. Therefore IPIi'1 = x + :. But IPFlI = e ( x + f ) = ex+a. Also l P L l = f - x andsince (PF21 = elPLl we have tha t (PF21= e ( % - x ) = a-ex . Adding both results we have that J P F l J + IPF2[ = 2a.


This property gives a simple method of drawing an ellipse. Fix two pins (the foci) on paper and pass a closed loop of thread round the pins. The length of the loop should be greater than the distance between the pins. Use a pencil or pen to pull the loop taut. Then trace out the locus of all points thus obtained.

If in the equation of an ellipse in standard form, we set a = b, that equation reduces to X 2 + y2 = a2 which is easily recognizable as the equation of a circle centre (0,O) in the XY-axes and radius equal to a . Also b2 = a2(1 - e2) becomes 1 = 1 - e2, so that e = 0. Hence a circle is an cllipse of eccentricity zero.

How is it physically possible to have eccentricity zero? Recall that e is the ratio8 of the distance to the focus and the distance to the directrix. To make this ratio vanishingly small it is sufficient to make the denominator large. This is achieved by putting the directrix far away (indeed the distance from the centre of an ellipse to the direcrtix is + and this becomes large as e -t 0). The effect of a positive eccentricity is to squash the circle invrards along the Y-axis (see Fig. 3.12). As e increases b becomes smaller till as e + 1 b - 0. In Fig. 3.13 0 < e l < e2 < 1 and therefore b2 < bl for the two ellipses shown.

ellipse (q > 01

3 ellipse (c2 z q

FIG. 3.13 Ellipse as squashed circle

3.3.4 The Hyperbola

By equations (8) to ( l l ) , the standard form of the equation of the hyperbola is

where b2 = a2(e2 - 1) and e > 1. (i) Y = 0 * X 2 = a2 , i.e., X = f a intercepts on X-axis.

X = 0 Y2 = -b2. Thus there is no intercept on the Y-axis. (ii) Replacing X by -X or Y by -Y does not alter the equation of the hyperbola. Therefore it is

symmetrical about both the X-axis and the Y-axis.

(iii) = $ - 1 > 0 3 x2 2 a2 i.e., 1x1 > a. Thus all points of the hyperbola lie outside the range -a < X < a .

(iv) As X, Y + km $, become large compared with 1. Thus = $ - 1 r; for large values of X . Hence % w f +. The two lines

92 Conic Sections

are called asymptotes of the hypcsrbola. Clearly since = - 1, then IY 1 < 1: 1 1x1 so that the'hyperbola lies between the asymptotes (See Figure 3.14).

FIG. 3.14 Sketch of hyperbola in standard form

(v) The directrix of the hyperbola is the y-axis and its equation is x = 0. In the new X Y coordin system this has the equation

Similarly the focus is a t the point (a(*), 0). This then becomes

= ae > a since e > 1 and Y = y = 0. e

By symmetry the hyperbola has two foci at (f a e , 0) and two directrices with equations X = fa (See Figure 3.14). Note that by conventior~ c is associated with the variable having the positive coefficient. The

foci also lie on this axis which is called the axis of the hyperbola. A point where the hyperbola intersects this axis is called a vertex of the hyperbola.

Example 9

Sketch the hyperbolae

X 2 Y 2 (i) - - T = 1 1

H 3

In each case find the coordinates of the foci and vertices and the equations of the asymptotes and directrices. . Solution

(i) a 2 = 3, b2 = $. Since b2 = a 2 ( e 2 - 1) we have that

Therefore

e = fi and ac = E= ifi.

3.3 Geometry and Sketches of the Cpnics 93

The foci and vertices are at (f ae, 0) = (&id, 0) and (fa, 0) = (f $, 0) respectively.

The asymptotes are Y = f t ~ , i.e., Y = &fix.

The directrices are the straight lines X = f: = f a = f-&. This hyperbola is similar to the one in Figure 3.14.

9 Here the Y-axis is the axis of the hyperbola. a2 = 4, b2 = 5 and therefore e2 = 1 + 5 = 1 + = a. Hence e = i. Since ae = 2 . = 3, the foci are at (0, f ae) = (0, f 3). The vertices are a t (0, fa) = (0, f2). The asymptotes of $ - $ = 1 are the equations Y = fix obtained by replacing 1 by zero in

the equation of the conic. Thus the asymptotes have the equations Y = f fix = &AX. The

directrices are perpendicular to the axis of the conic at a distance f = & = 2 x 3 = $. Hence the equations of the directrices are Y = f!. Figure 3.15 illustrates this hyperbola which has the Y-axis as its axis.

FIG. 3.15 Hyperbola 7 - $ = 1

The hyperbola also has the property that it is the locus of all points P such that the magnitude of the difference between the distances from the two foci Fl and F2 is a constant. We now prove this.

Example 10

Let P ( z , y) be any point on the hyperbola $ - $ = 1 with its foci'at Fl(-ae,O) and F2(ae,0). (see Figure 3.16). Then IIPFl I - IPFzII = 'L ' l r

FIG. 3.16

94 Conic Sections - _ _.- _--r -

Let K L P be the line through- P parallell6 the x-axis (Figure 3.16). The coordinates af K,L are (-$, y) and (%, y) respectively. Therefore IPK) = x + :. But IPFII = elPK1 since P lies on the hyperbola. Thus IPFl( = e ( x + f ) = ex+a . Similarly lPL( = x- t and IPF2( = e(PLI = e(x- t ) = ex - a. Thus IPFll - (PF2J = 2a. By taking P on the left hand arm of the hyperbola, we similarly obtain I PF2( - IPFl I = 20. Therefore I(PFl I - JPF2JI = 2a.

Ro

FIG. 3.17

This property can be used to construct the hyperbola. Take two fixed points (the foci) a distance 2a + 2ro apart where a , r o are arbitrarily chosen (see Figure 3.17). With F2 as centre and radius ro draw a circle Ro to cut F1F2 at A. The circle Lo, centre Fl and radius ro + 2a also cuts FI F2 at A. Draw other pairs of circles Li, &, i = 1,2,3, . . . centre Fl and F2 respectively such that each pair of radii differ by 2a as for Lo and Ro. Mark the two points of intersection of Li and Ri. Join these with a smooth curve t o obtain the right hand arm of the hyperbola. We may construct some of the points on the left arm of the hyperbola in a similar fashion.

The rectangular hyperbola

The standard equation of a hyperbola

with b2 = a2(e2 - 1) and e > 1 reduces

(17)

in the special case of b = a to the equation

xa - y2 = a2,

The eccentricity of this hyperbola is thus given by 1 = e2 - 1 so that e = fi. The gradient of its asymptotes are ,& = f 1 (by equation (16)) so that each asymptote is inclined a t an angle of 4 5 O to the X-axis (see Figure 3.18a).

a. In X-Y coordinates FIG. 3.18 A rectangular hyperbola

b. Asymptotes as axes


Since the asymptotes are perpendicular to each other, this hyperbola is called a rectangular hyperbola. Note that its equation x2 - y2 ='a2 is factorizable to obtain ( X - X ) ( X + Y) = aa, If we use the asymptotes as rectangular axes by putting X - Y .= < and X + Y ='q in (17) we obtain

as another standard form of the rectangular hyperbola. Figure 3.18b il1ustrates.a sketch of a rectangular hyperbola using its asymptotes as axes. Figure 3.18b is what would be obtained by rotating Figure 3.18a. through 4 5 O anticlockwise.

Equation (17) deals with only one possible hyperbola, that in which the axis of the hyperbola is the X-axis. However, it is possible to have $ - = 1 so that the Y-axis is the axis of the hyperbola. This latter case leads to rectangular hyperbola ( q = -a2 if we set X -Y = (, X +Y = q as before. Therefore the most general form of the rectangular hyperbola must be

where c is a real constant which may be positive, negative or aero. Figure 3.18b illqtrates the case c > 0. If c < 0 then ( and q must be opposite in signs and the hyperbola lies in the second and fourth quadrants as shown in Fikure 3.19

4

FIG. 3.19 T ~ I P hyperbola :'ty - d 2- ,

If E = 0, the hyperbola b e e o m ( q = 0 in which case *t least one of f , q muat be item. The rmtmgulat hyperbola in such a case degenerates to the pait of lines given by 4 = 0 m q = 0'.

Combining all the three types of conics we have the following results for a conic in standard form: (i) the focus is on the axis (or 011 major-axis if the conic has more than one axis) at a distance ae

from the origin. (ii) the directrix is perpendicular to the (major) axis at a distance from the origin.

a2(1 - e2) for e < 1 (iii) b2 = { a ( e - 2 ) for e > 1

3.3.5 Working i n the original coordinates

Given the equation of a conic we have shown how to reduce it to the standard form provided that its cquation has no term involving the crossproduct xy. (The interested reader may refer to Chapter 4, Section 3 lo r techniques of getting rid of the crossproduct term when it is present as in Example 1 of Section 4.2.) A conic once reduced to its standard form can be sketched and its characteristics like focus, directrix, ( ccc,ntricity, asymptotes etc identified in terms of the original coordinate system. T h ~ s is best illust rated by an example.

96 Conic Sections

Example 11 In each of the following cases, identify • he conic and determine the coordinates of its foci, centrtb and vertices as applicable, Also obtain the equations of the directrices and asymptotes, if any:

(i) 4 ( ~ + 5 ) ~ - 3 y + 2 = 0

(x - + y + 2)2 (ii) - - - 9 5

- 1 (x - (y - 4)2

(iii) --- 4 8

- 1.

Solution

(i) 4(x + 5)2 - 3y + 2 = 0. Therefore

2 3 2 4 ( ~ + 5 ) ' = 3 y - 2 = 3 ( y - - ) and (x+5) '= ;(y- ?)

3

3 i;e., x2 = 4aY where X = x + 5, Y = y - and 4a ri + a = > 0. I I i ~ s i s the equation of a parabola with y-axis as axia. Its vertex is a t X = U;Y = 0. That is, x + 5 = 0 r = -5 and

2 y - 5 = 0 + y = 8. Thus the vertex is at the point (-5, $1. We sketch the parabola in terms of parallel axes through the centre (see Figure 3.20). Note that Y = y - 5 2 0. The focus is a t (-5, + a) = (-5, $ + &) = (-5 41) 48 '

The directrix is at Y = -a i.e.

(x - 4)' + y + 2)2 X 2 Y2 - I = - + - - = I . (ii) - - - 9 5 9 5

The centre of the ellipse is at X = 0,Y = 0, i.e. a t x - 4 = O,y+2 = 0 + x = 4,y = -2. Thus centre = (4, -2). The major axis is the X-axis and is thus parallel to the x-axis where X , 1' are parallel axes through the cent r

-10 -8 -6 j -4 -2 2 iSXx

3 2 FIG. 3.20 ( x + 5)' = ,(y - -) 3 We again sketch the ellipse in terms of X and Y (see Figure 3.20). Note that CD = 4 > a = 3 and CE = 2 < b = 6 so that the ellipse does not cut the y-axis wheras it cuts the x-axis a t two points


Thus e = $ and ae = 3 . $ = 2. Similarly fi = 3 . 3 - 9 e . 2 - 2 ' The foci of the ellipse are at (4 f ae, -2) = (4 f 2, -2) = (2, -2) and (6, -2). The vertices are at (4 f a, -2) = (4 f 3, -2) = (7, -2) and (1, -2) The equations of the directrices are X = f % = f i.e.

(x - ;)2 (?,-4)2 (iii) --- X 2 Y2 - I = - - -

4 8 a2 b2 = 1

where a2 = 4, b2 = 8 and e2 = 1 + = 1 + = 1 + 2 = 3. Thus e = fi and ae = 2 . a. The centre is a t the point (+, 4). The axis of the hyperbola is parallel t o the x-axis since the X-axis is the axis of the hyperbola. Sketching in terms of parallel axes through the centre we obtain Figure 3.21. CD = < a = 2 and so the hyperbola does not cut the y-axis.

(x - f ) ? (y - 4)2 FIG. 3.21 --- 4 8

- 1

The asymptotes are at Y =' f ! ~ = &fix = fix. These are the equations y - 4 = f f i ( x - f ) .

The foci are a t (3 f ae, 4) = ($ + 2 8 , 4 ) . The vertices of the hyperbola are each a distance a from the centre along the axis. Hence these have coordinates ( 3 k a , 4) = (4 f 2,4) = ($,4) and (-%, 4). These are marked V2, VI respectively in Figure 3.21.

Exercises 3.3

. Sketch the following conics. In each case indicate clearly the coordinates of the foci, and the equations of its directrices and asymptotes, if any.

98 Conic Sections

Sketch the following conics. In each case detcrmine the coordinates of its centre, foci, vertex. AIso obtain the equations of its directrices and asymptotes as applicable.

3.4 Tangents and Normals

Tangents and normals are first and foremost straight lines. Recall (from Chapter 1) that essentially two bits of information are required before one can derive the equation of any straight line: these are

(i) the gradient of the line, and (ii) the coordinates of a point on the line. In solving a problem we must therefore try to obtain

these two pieces of information.

3.4.1 Tangent a n d no rma l a t a point o n any curve

. . \

FIG. 3.22 Tangent and normal to a curve

In deriving the equation of the tangent or normal to any curve at a specified point on the curve, the second of the two requirements mentioned above, namely, the coordinates of the point a t which the tangent/normal is required, is already provided. Thus the problem reduces to finding the gradient of the tangent or normal. However, since the normal and tangent at any point of a curve are perpendicular to each other, (see Figure 3.22) it is sufficient to determine the gradient of the tangent. The gradient of the normal is then easily obtained by taking the reciprocal of the negative of the gradient of the tangent.

How can we find the gradient of the tangent to a curve at a given point of the curve? Of course, if the equation of the curve is given in the form y = f (x), the gradient is 2 evaluated a t

3.4 Tangents and Normals 99

the required point. But often y is not given explicitly but implicitly as g(x, Y) = 0. Example of such equations are

and

As we can see from (20) it is often difficult or impossible to solve for y explicitly in terms of x. Fortunately, we need not solve for y. Since y is a function of x , we may differentiate the equation aa it stands implicitly with respect to z. Recall that if u(y) is a function of the variable y, where y is some function of x , then by the chain rule

du du dy -- _ - . - dx dy dx'

Thus differentiating (20a) term by term yields

Cancelling out the common factor of two and simplifying, we obtain

as the gradient of the tangent at any point (x, y) on the hyperbola (20a). Note that equation (20a) is the solution of Example 1 of section 3.2.1 and its eccentricity e = fi > 1. For example, at the point (2,3), the gradient of the tangent is therefore given by

Hence the equation of the tangent to the hyperbola (20a) a t (2,3) is

The normal at the same point has gradient -$ = 5 and eq~at~ion

Remark 3.2

Equations (22) and (23) illustrate a common property of perpendicular lines, namely,

are perpendicular lines. Their gradients are respectively -f and and thus have a product equal to -1. The value of the constant k may be found by substituting the coordinates of a point on the line.

100 Conic Sections

Exercises 3.4.1

Find the equations of the tangent and normal to the following drves at the given point.

1 ex + x ) + 4 x y = 1 - y at ( 0 , O ) . 2 x y = 4 at ( 4 , l ) .

11 5 13x2 + 4 8 x y + 27y2 - 166x - 188y + 313 = 0 at (2, -).

27 6 9 ~ ~ + 4 ~ ~ + 1 8 ~ - 2 4 y + 9 = 0 at ( l , 3 ) .

By implicit differentiation obtain 2 for the following equations and thus find the equations of the tangent and normal to the curves at the given point.

7 In(x + y ) + cos x + y = 2 at ( 0 , l ) .

3.4.2 Tangent and normal at a point on a conic

The methods developed in Section 3.4.1 are enough for finding the equations of the tangent and normal to any curve at a point on the curve. We recommend its use. However, by applying the same process to an arbitrary conic, we can obtain a general formula which is'applicable to all conic sections. Although the use of such formulae is unnecessary we include them here for the benefit of those who might prefer them to working from first principles. We shall derive these results initially for each conic and finally for the general conic.

Example 12

Find the equations of the tangent and normal to the parabola y2 = 4ax at the point ( x l , y l ) on the parabola.

Solution

Differentiate the equation y2 = 4ax implicitly with respect to x to obtain

Hence the equation of the tangent a t ( X I , y l ) is

Cross multiplying and collecting terms we have 1 I

yy , = 2ax + yT - 2axl = 2ax + 4axl - 2axl since ( x l , y l ) lies on cur*

= 2ax + 2ax1 = 2a(x + XI)


that is

Comparing (25) with the equations of the parabola we see that the y2 term goes into product yyl whereas the x goes into i ( z + 2,).

The gradient of the normal is dy Y l - _ -- dx 2a

(negative of the reciprocal of gradient of tangent). Hence the equation of the normal is

which simplifies to 2ay + ylx = yl(2a + XI).

In the alternative we could rearrange (25) and use (24). Thus (25) gives yyl - 2ax = 2axl. Hence by (24) the equation of the normal is 2ay + ylx = k where k is evaluated by substituting ( X I , yl) into the equation for k. However only the formula for tangent need be remembered.

For example, the equation of the tangent to the parabola y2 = i x at (12,3) is given by yy1 = . g(x + xl) or 3y = :(x + 12). This simplifies to 8y = x + 12. The gadient of the tangent is t h u t i :from 8y = x + 12) and hence that of the normal is -8. Therefore the equation of the normal is y - 3 = -8(x - 12) or y + 82 = 99. Note that by applying (24) we have equation of the tangent is 8y - x = 12. Hence equation of normal is y + 82 = k = l(3) + 8(12) = 99 as before.

Example 13

Find the equation of the tangent to the ellipse 5 + 6 = 1 at the point (xl, yl) on the conic.

Solution

To find the gradient of the tangent we differentiate the equation of the conic implicitly with respect to x, This yields

Hence at (XI, yl) the gradient of the tangent is -52 and its equation is

Rearranging we have ..

since (21, yl) lies on the ellipse. Thus the equation of the tangent to the ellipse 5 + = 1 st the point (XI, yl) is

102 Conic Sections

Once again squared terms become products as x2 becomes 1x1, y2 becomes y y l .

z2 Y 2 For the ellipse - + - = 1 a t ( 2 , 3 ) we therefore have that the equation of the tangent is

16 12

The gradient of the tangent is thus -3 and so that of the normal is 2. The equation of the normal is then

y - 3 = 2 ( x - 2 ) or y - 2 x + 1 = 0 .

We could, of course, use ( 24 ) to write down the equation of the normal as 2 x - y = k = 2 ( 2 ) - l ( 3 ) = 1, i.e. 22 - y = 1 as before.

Example 14

Find the equation of the tangent to the hyperbola $ - = 1 at the point ( z l , y l ) on the conic.

Solution

Differentiate the equation of the conic implicitly with respect to x .

2x - - -- d y b2 x d y = 0 which implies - = --

a2 b2 dx dx a 2 y '

Hence the equation of the tangent at ( z l , yl) on the hyperbola is

Rearranging we have

Therefore

( 2 7 )

In numerical illustration let us find the equations of the tangent and normal to the hyperbola y2 x2 ----

4918 49/16 - 1 at the point ( z , i).

he equation of the tangent is

that is,

that is, 4 y - 42 = 7

3.4 Tangents and Normals 103 --

The gradient of the tangent is thus 4-1 so t h a ~ _ t h e m ~ i s - 1 . Consequently, the equation 7 of the normal is y - = - l ( x - T ) or y + x = +. Multiplying out gives 4 y +& = 21.

Example 15

Obtain the equation of the tangent to the conic ax2 + bxy + cy2 + 292 + 2 f y + d = 0 a t the point ( X I , y l ) on the conic. -- -- . Solution - - - -

We differentiate the equation of the conic implicitly with respect to x to obtain the gradient of the - - ,li"-C---C---- - tangent.

Therefore d y 2ax + by + 29 - = -

. , d x b x + 2 c y + 2 f '

The equation of the tangent a t ( x l , y l ) is

-. & --

Cross multiplying and collecting,terms, we have

l.e., 2axx l + b(Yxi + x y i ) + 2 c w l + 292 + 2 f y = 2ax: + 2bxly l + s ~ ~ T + 2gx1 + 2 f yl

.- - - - -2(2gx1 + 2 f yl + d ) + 29x1 + 2 fyl---- - . -- -. -- '

since ( X I , y l ) lies on the conic = -29x1 - 2 f y l - 2d.

Therefore 1

2axx l + 5 b ( ~ ~ ~ + W l ) + 2Wy1 + 2g(x + x i ) + 2 f ( y + y l ) + 2d = 0

Comparing this equation with the equation of the conic we observe that

1 x -r i ( x + X I ) and

1 v - +Y + ~ l )

104 Conic Sections

where the arrow -, means is "replaced by".

Each of the previous results (25 ) to (27 ) for the special conics in standard form agrees with this more general result. Indeed recall that for the circle (an ellipse of eccentricity zero-see Section 3.3.3) x 2 + y2 + 292 + 2 f y + d = 0 , the equation of its tangent at ( x l , y l ) is given by 2x1 + yyl + g ( x + X I ) + f ( y + y l ) + d = 0 and this also agrees with (28) .

To illustrate this result we go back to the conic of equation (20a) , namely,

What is the equation of the tangent t o this conic at the point (2 ,3 )? By (28 ) this tangent has equation

13x21 + 2 4 ( W + y z l ) + 2 7 ~ 1 - 83( x + 21) - 94(y + y l ) + 313 = 0

Thus 262 + 7 2 2 + 48y + 81y - 832 - 166 - 94y - 282 + 313 = 0

as before.

Exercises 3.4.2

Find the equation of the tangent and normal to each of the following conics a t the specified point.

1 x y = -4 at ( 4 , -1) '

2 4y2 - 32 - 8 y + 10 = 0 at (T, -1)

3 x 2 = 8 y at ( 4 , 2 )

4 13x2 + 48xy + 27y2 - 1662 - 188y + 313 = 0 at (-&, 3 )

3.4.3 Tangents from a point

In this type of problem we are given a curve, say, f ( x , y ) = 0 and a point A not on the curve (see Figure 3.23). We are then required to find the equation of a line which passes through the given point A and is tangent to the curve at an unspecified point. The line APB in Figure 3.23 is one such


tangent whereas :1QC is not. We therefore need to determine what the gradient of a line through A must be for such a line to be tangent to the 'curve at some point.

FIG. 3.23 Intersection of lines through A with curve.

Example 16 Find the equations of the tangents from the given points to the following conics. For each such tangent, determine also the coordinates of the point at which it touches the conic.

x2 (i) - - - y2 = 1 from the point (3,lO).

2 25 x2 y2

(ii) - + - = 1 from the point (-12,l). 16 3

Solution

(i) Let the gradient of a line through the point (3,lO) be denoted by m. The equation of the line is then y - 10 = m(x - 3) that is,

y = m x + 1 0 - 3 m . (1)

Substituting for y from (1) into the equation of the hyperbola, we have

Equation (2) is a quadratic in x, the coordinate of the point where the line (1) intersects the hyperbola. For the line to be a tangent, it must touch the hyperbola. Hence (2) must have only one solution (that is, two coincident roots). This is true only when the discriminant is zero; that is,

Collecting terms in powers of m, we get

106 Conic Sections . '

Therefore m = or 5

Substituting these values into ( 1 ) we obtain

and

for m = 5 and 9 respectively. To find the coordinates of the point of tangency, let th 's be the point ( 2 , y ) . A quadratic

equation a x 2 + bx + c = 0 with coincident roots has solution x = -&. Hence from (2), Z is given by

For m = 5 , 5 = 2 ( 5 ) ( i o - i g j

= 2. 25 - 50

To find the corresponding values of jj we note that ( 2 , y ) lies on the tangent with equation ( 3 ) . Thus for m = q, x = 10,7y = 25(10) - 5 - y = 35. For m = 5 , 2 = 2 and jj = 5 ( 2 ) - 5 = 5. Therefore the points of tangency are ( 1 0 , 3 5 ) and ( 2 , 5 ) respectively.

x L yL (ii) Find the equations of the tangent to - + - = 1 from the point (-12,l).

16 3 Let the gradient of the tangent be m. Therefore, the equation of the tangent is y - 1 = m ( z + 12)

Substitute ( 4 ) into the equation of the ellipse to find the points of intersection

x2 ( m x + 12m + 1)' - + = 1 16 3

We have coincident roots if and only if

After dividing throughout by the common factor 64, multiplying out and collecting terms in de- scending powers of m, we obtain,


Once again the m4, m3 terms cancel out to give

Therefore

Thus the tangents are 1 12 1 12

y = - x + - + 1 and y = - - x - - + 1 16 16 4 4

where we have substitutied for m in (4). These simplify to

and

LTo find the points of tangency (5, y) we use (5) and (4).

Therefore

1 - 3 m = - - , z = - 2 = j 4 j + ( - 2 ) + 8 = 0 * 5 = - - 4 2 '

Thus the coordinates of the points of tangency are (-g, E) and (-2, -%) respectively.

Remark 3.3

(1) The quadratic equation (2) or (5) for the x-coordinates of the points of intersection of the line and the conic contains the unknown gradient m. Indeed, in the notation of the standard quadratic ax2 + bx + c = 0, the coefficients a , b, c of that equation are all of degree 2 in m. Thus whell we impose the condition b2 - 4ac = 0 for tangency, we expect a polynomial of degree four in m. Such a polynomial would in general be difficult to solve. Fortunately, in the two examples above the m4 and m3 terms cancelled out thus leaving us with a quadratic in m which is immediately solvable! We shall show in the next example that this happy condition will always occur for any conic section.

(2) Since the equation for in is a quadratic, there are at most two tangents from a point not on a conic to that conic. Can you think of situations where there is only one tangent or none at all from a given point to a conic? See Exercise 3.4.3 question 5. ,

108 Conic Sections

Example 17

Show that, except for one case, the condition for tangency always yields a quadratic equation in the gradient, m , of the tangent to a conic from a point not on the conic.

Solution

Let the conic be a x 2 + b x y + c y 2 + 2 g x + 2 f y + d = 0

and let the point not on the conic be (21, y l ) . The equation of a line through ( z l , y l ) with gradient m is

y - y l = m ( z - z l ) or y = m x + y l - m z l . (8)

Substitute this last equation into the equation for the conic

ax2 + bx[mx + yl - m x l ] + c [ m x + yl - mx1I2 + 292 + 2 f [ m x + yl - m z l ] + d = 0

Write each coefficient in powers of m and obtain

This equation has coincident roots if and only if its discriminant is zero; that is,

or equivalently ~rn'+Qm~+Rd+Snr+T=O

where P = 4c2x; - 4c2x; = 0

& = -4cx1 ( 2 f - bzl + 2 ~ 9 1 ) - 4bcz: + 4c(2 f z l + 2x1 ylc) = 0 .

Hence in all cases, the coefficients of m4 and m 3 vanish. Thus except when R = 0 , equation (10) is quadratic in m.

Exercises 3.4.3

In each of the problems 1-8, (a) find the equations of the tangents from the given point to the given conic; (b) determine also the point at which the tangent touches the conic. Explain any case where there is no tangent.

x2 1 ---- y2 - 1 from the point (0 ,5 ) .

2 25 x2 y2

2 - + - = 1 from the point ( 4 , -1). 16 3

3.5 Other forms of the equation of a conic 109

x2 y2 5 - + - = 1 from (1, l) . What is the position of the point (1, l ) in relation to the ellipse?

4 9 6 x 2 + 8 y = O f r o m ( ~ , - 3 ) .

7 y2=8xfrom(1,1) .

8 xy = -10 from (6,5).

9 Find the condition on c for the line y = rnx + c to be tangent to the parabola y2 = 4ax for a given rn.

3.5 Other forms of the equation of a conic

So far 'we have been describing conic sections in terms of Cartesian coordinate axes. In this section we shall study two other methods of describing plane curves. These are the parametric and polar equations.

3.5.1 Parametric equations of the conic

Suppose that a particle moves in the x-y plane and we keep track of its positions over some time interval I = { t 1 a 5 t 5 p ) . Then clearly, x = f(t) and y = g(t) for some functions f , g and 1 E I.However, the particle traces out a path or curve as it moves. For example, x = 2 coswt, y = 2sinwl trace out the circle x2 + y2 = 4. The equations x = f(t), y = g(t) are called the parametric equations of the curve generated by the moving point and t is called the parameter. Tbus the parametric equations are the natural way to describe the position of a moving point. Although the idea of parametrization of curves originated with time t as the parameter, the method has grown and any convenient variable may now be used as the parameter. Consider the following example.

Example 18

A car tyre of radiue a is pierced by a nail which'sticks to the tyre. Find the parametric equations of the curve traced out by this nail as the car moves in a straight line.

Solution Let the nail be picked up at the point 0 which we take as the origin of the x-y coordinate system. After a time interval 1, let the nail be at position P(x, y) (see Figure 3.24) where the radius vector C P from the centre of the tyre to the nail has turned through an angle 0 from its initial position. Now IOA) = length of arc AP. = a0 since the tyre rolls on the road. Thus the coordinates of C , the centre of the tyre are (a0, a). Also

and

110 Conic Sectioris

0 R A

FIG. 3.24

Thus the point P(x, y) lies on the curve whose parametric equations are . ,

x=a(B-sing), y=a(l-cosO), 0 2 0 .

0

Note that in'the above example it is not easy to eliminate 0 and obtain the Cartesian equation of the curve, called a cycloid. Fortunately the conic sections in standard form are easy to parameterize. For instance the ellipse $ + 6 = 1 suggest8 the use of sin2 0 + eos2 0 = 1 whereas the hyperbola

$ - = 1 suggests sec2 t9 - tan2 t9 = 1. The usual parametric equations for the conic sections in standard form are as follows:

CONIC STANDARD FORM PARAMETRIC EQUATIONS

Hyperbola (rectangular) xy = c2

Parabola I . y2 = 4ax z = ata, y ='2at

Note that the parameter for the ellipse is the angle 0 of plane polar coordinates, Figure 9.25 illustrates the relationship between the ellipse $ + $ = 1 and the two circles x2 + y2 = a2 and x2 + y2 = b2. The y-coordinate of P is given by that of S whereas the x-coordinate of P and Q are the same.

I

FIG. 3.25 An ellipse $ + 6 = 1 between the circles of radii a and b.

3.5 Other forms of the equation of a conic 11 1

Example 19

Give parametric equations for the following conics:

(i) 16(x - 3)' + 25(y - 2)2 = 400 (ii) 4x2 - 82 - Y2 + 6y - 7 = 0 (iii) 3x2 = 1 - 2~ (iv) (x + l)(y - 2) = 9.

Solution (x-3)2 (y-2)2

(i) 16(x - 3)2 + 25(y - 2)' = 400 * - + - - 25 16

- 1.

Therefore,we must put x - 3 = 5cos0 and y - 2 = 4sinB

so as to get cos2 6 + sin2 6 = 1. Tllus parametric equations for (i) are

2 (ii) 42' - 82 - y + 6y - 7 = 0 4(z2 - 22) - (y2 - 6y) = 7, Completing squares we get

Therefore (x - 1)2 (y - 3)2 ----

1 2 - 1. -

2

Comparing this with the identity sec2 6 - tan2 0 = 1

we set

Here 4a = -:, so that a = - i. Therefore

Thus parametric equations are z = - i t , y = - l t 2 . 6

(iv) (x + l)(y - 2) = 9. This is a rectangular hyperbola of the form xy = c2 with c = 3. We see from the table above that the appropriate parametric equal :ens are

112 Conic Sections

Example 20

Find the Cartesian equation of the curve which has the parametric equations x = 2 + cost, y = 1 - cos 2. Identity the curve.

Solution

y = 1 -cos2t = 1 - (2cos2t -1) = 2 - 2 c o s 2 t = 2 - 2 ( ~ - 2 ) ~ s incex -2=cos t .

Hence y - 2 = -2(x - 2)2. This is a parabola.

Exercises 3.5.1 Obtain parametric equations of the following conics.

Find the Cartesian equations for the following curves whose equations are given in parametric form. Identify each curve.

10 x = 1 - c o s t , y = 2 + s i n t 11 x = s i n t l y = c o s 2 t

12 x = 2cos8, y = 1 + cos20 1 13 ~ = 2 - ~ c o s t , y = 3 + $ s i n t

14 x=&, y = t + 2 15 x = -2e-', y = 3et

3.5.2 Working with parametric equations

Since the parametric equations of the respective conics have only one variable, namely, the parameter, any Cartesian equation in x and y can easily be reduced t o one in the single parameter, Therefore the parametric equations are often useful in solving problems on intersection or locus. However our relative inability to solve equations in the respective parameters (especially those involving trigonometric functions) often curtail our full use of this technique.

Example 21

Find the points of intersection of the straight line y = 3x + 2 with the curve whose parametric equations are x = 2t2 - 2, y = 6 - 41.

Solution

Given the line y = 3x + 2, the point (2, y) will also lie on the curve if they satisfy the parametric equations. Hence 6 - 4t = 3(2t2 - 2) + 2 i.e.

Factorising we have (32 + 5)(t - 1) = 0. Thus t = -; or 1. Hence the coordinates of the points of intersection of the line and the curve are ( y , F) for t = -; and (0,2) for t = 1.


Example 22

Find the equations of the tangents from the point (1 , l ) to the curve y2 = 8x [Problem 7 of Exercises 3.4.31

Solution The parabola y2 = 82 is equivqlent to y2 = 4ax where a = 2. Therefore its parametric form is x = at2 = 2t2 and y = 2at = 42. ~ e t the gradient of the line through the point (1 , l ) be m. Its equation is thus y - 1 = m(x - 1). This intercepts the parabola where z = 2t2, y = 4t and thus 41 - 1 = m(2t2 - 1) i.e. 2mt2 - 4t + 1 - m = 0.

The line is a tangent when we have coincident roots for this quadratic in t , i.e.

We now show how to find the equations of the tangents and normals to various conics by using their parametric equations.

If the parameter is u, so that x = x(u), y = y(u) are functions of u, then

a. Ellipse: x = a cos 0, y = b sin 0

for apy point with parameter 0 on the ellipse. Ttcrefore equation of tangent to the ellipse a t the point el is

b cmOl y - bsinOl = ---(x - acosel)

a sin el

Multiplying by a sin 81, yields

a sin y - ab sin2 el = - b cos el x + ab cos2 el

which simplifies to

(29) bcosO1 x+asinO1 y = ab.

Remark 3.4

(1) In deriving equation (29), 01 is an arbitrary but fixed point, just like (xl, yl) and not a variable; the variables are x and y which give the equation of the tangent.

2x1 Y Y l (2) Recall the equation (26) - + = 1 for the tangent a t the point (q, y l ) If we substitute a2

x1 = a cos e l , yl = bsin dl into this equation we obtain equation (29). We have, however, shown how to derive equation (39) independently on its own.

114 Conic Sections

a sin t9 The gradient of the normal is +- so that the equation of the normal a t el is

bc-e

a sin 81 y - bsin el = -(x - a cos el)

b COS el That is

b COB el - b2 sin el cos el = a sin dl x - aa sin cos el which simplifies to

(30) a sin el x - b cos O1 y = a2e2 sin 61 cos 81.

b. (i) General hyperbola

Here x = a sec 0, r/ = b tan 0 so that

dx dy - = a sec 0 tan 8, - = b sec2 0. do do

The gradient of the tangent ie

Therefore the equation of the tangent a t O1 is

b 1 y-btanO1 = --(x-asecO1)

a sin el

sin2 el asinO1 y - ab- = bx - absecel

cos el

ab = -(I - sin2 81) = ab cos dl. cos el

Thus

The equation of the qormal is

i.e. by - b2 tan2 el = -a sin el x + a2 tan el or a sin O1z + by = (a2 + b2) tan el = a2e2 tan 81. Thus

(32) asinel x + b y = a2e2tand1.

b. (ii) Rectangular hyperbola

C x = d , y = -

t '

3.5 Other forms of the equation o f a conic 115

Therefore

Thus the equation of the tangent at t l is

The normal has gradient t : and thus has the equation y - = t : ( x - ctl) i.e

c. Pa rabo la

The parabola y2 = 4ax has the parametric equations x = a t 2 , y = 2at. Therefore, its gradient

Therefore the equation of the tangent a t the point t 1 is y - 2atl = k ( x - at: ) or

The equation of the normal is accordingly y - 2atl = - t l ( x - a t ; ) or

(36) y + t l x = 2atl + at:.

Example 23 Show that the tangents a t the ends of a focal chord of the parabola y2 = 4ax intersect a t right angles to each other.

Solut ion

FIG. 3.26 A focal chord

..

116 Conic Sections

A focal chord (see Fig. 3.26) is any chord of the parabola which passes through the focus a t (a, 0) in x-y coordinates. Let PFR be such a chord with gradient m. The equation of P F R is

If this line intersects the parabola y2 = 4ax a t the point t , then x = at2, y = 2at so that 2at = mat2 - ma or

mt2 -2t -rn = 0 (ii)

with

Denote these by t l and t z . The equation of a tangent at any point t of the parabola is by (35) y - 4 x = at and this has gradient f . The tangents at t l , t2 will be perpendicular if the product of their gradients is -1. Now

1 1 1 - . - = - - - from (ii) t l t2 t1t2 -m/m

Thus these tangents are at right angles to each other.

Exercises 3.5.2

1 Find where the straight line y = 3x + 4 cuts the curve whose parametric equations are 2 4

(i) z = 2 t , y = - (ii) x + 2 = 2 t 2 , y - 6 = - 2 t (iii) x = -cosO, y=4s inO t 3

2 Solve problems 2, 3, 5, 6, and 8 of Exercises 3.4.3 using parametric representation. Hint for problem 2: sin 4 = a has only one solution iff 4 = $ (for a > 0) or -; (for a < 0) a2 = 1 where 0 = a(rn).

Find the equations of the tangents and normals to the conics at the specified points in problems 3-10.

lr x = sint + cost, y = sint - cost a t t = -

3



Obtain the equation of the conic satisfying the conditions in problems 1-3.

1 Directrix x = 3 , focus ( O , l ) , eccentricity e = 5. 2 Directrix x + y = 6 , focus ( - 2 , 3 ) , eccentricity e = 5 . 3 e = 1, focus lies on the line y = 1 ; the directrix is parallel to the y-axis while the conic passes

through the points ( - 1 , l ) and (4,2).

Put the conics in problems 4-6 into standard form and determine the type of each conic. Find also the coordi~~ates of the centre or vertex as appropriate.

4 ~ ~ + 4 ~ ~ - 2 x + 4 ~ + 1 = 0

5 y 2 + 3 x - 2 y = o

Sketch the conics in problems 7-9. In each case determine the coordinates of its centre, foci, vertex. Also obtain the equations of its directrices and asymptotes as applicable.

Find the equations of the tangent and normal to the curves in problems 10-11 at the given point.

10 9 x 2 + 4 y 2 + 1 8 x - 2 4 y + 9 = O a t ( -1 ,O)

11 ~ ~ + 2 x y - 3 ~ ~ = 1 a t ( l , O )

12 a. Find the equations of the tangents from ( - 1 , - 1 ) to the conic x2 = 8 y . b. Determine also the point a t which the tangent touches the conic.

x2 y2 13 Show that the lines y = m x f d w are tangent to. the ellipse - + - = 1 for all m.

a2 62 14 Obtain parametric equations of the conic 2xy - 3y + 42 + 1 = 0

15 Find the Cartesian equations of the curve whose equations in parametric form are x = sint + cost, y = sin t - cost. Identify the curve.

16 Find the equations of t h ~ tangent and normal to the conic x = -2e-', y = 3et at t = 0 .

Applications and General Equations of the Conics

4.1 Some Applications of the Conic

In this section we shall study some of the properties of the conics which have given rise t.o practical applications. These applications will also be described.

4.1.1 Parabola

The major characteristic of the parabola is its reflecting property: a small light bulb placed at the focus of the parabola produces a beam of parallel rays of light.

FIG. 4.1

Let P with parametric co-ordinates ( a t 2 , 2 a t ) be an arbitrary point on the parabola y2 = 4ax . Let the tangent TPQ to the parabola at the point P cut the x-axis at Q (see Figure 4 .1 ) . Through P draw the line P R parallel to the x-axis.

Example 1

Show that the lines PF and P R of Figure 4.1 make equal angles with the tangent TQP,

4.1 Some J ~plications of the Conic 119

Solut ion

The two angles marked P in Figure 4.1 are 'equal since P R 11 QFx . We i k d show that L F Q P = L F P Q which implies that a = 3. Recall that the gradient of the tangent to the prrabola at +.he point t is

So that the eq~~a t ion of the tangent at t i-

This simplifies to

1 y = -x + at (see also equation (35) of section 3.5).

t

At Q, y = 0 so that x = -at? Hence lQFl = a + (1t2 = a ( l + t2). Also

Therefore P = L F Q P = L F P Q = a, base angles of isosceles triangles

Remark 4.1

1 PF and P R are also equally inclined t o the normal a t P . 2 The above problem can be solved as easily by employing Cartesian ceordinates throughout. The equation of the tangent at (x l , yl) on the parabola y2 = 4ax is given by yyl = 2a(x + XI) (see equation (25) of section 3.4.2). At the point Q on the x-axis, y = 0 so that x + xl = 0 3 x = -XI. Hence IF&[ = a + x l . Also IPFI2 = (xl - a)2 + (yl - 0)2 = t: - 2ax1 + a2 + yf or IPFI2 =

2 x: - 2axl + a2 + 4ax1 = x1 + 2ax1 + a2 = (xl + a)2. Thus lPFl = xl + a = IFQI as for parametric methods. 3 An alternative method of solution is to verify that the gradient of PF is twice that of the tangent. Thus a + /3 = 2/3 a = P (see Exercise 4.1, question 5).

The reflecting property leads to perhaps the most common practical application of the conics in real life. By a well-known law of Physics, when light or sound strikes an object, part of that light or sound is reflected and in such a way that the incoming ray and the outgoing one make equal angles with the normal a t the points of impact. If the reflector is shaped like a parabola and the light source is placed at its focus, F , then by the reflecting property of parabola, all rays of light will come out parallel to the axis of the parabola (Figure 4.2(a)). On the other hand if rays of light parallel t o the axis of the parabola strike a parabolic mirror, they all will be reflected to the focus

120 Applications and General Equations of the Conics

(Figure 4.2(b)).

(a) 11ays from F FIG. 2

The first result is used whenever a strong beam of light is needed to travel far as in a soarch light, car head lamp, or even an ordinary torch light. The second result is the principle bt.l~ind the use of parabolic telescopes and TV/radio antennas to receive incoming signals. Focusing th- -e a t a point concentrates and amplifies the signal strength.

4.1.2 Ellipse

The ellipse also possesses a reflecting property.

Example 2

Show that the focal chords to any point P on the ellipse are equally inclined to the tangent or normal.

Solution Consider the ellipse

Let Fl (-ae. O), and F?(cte, 0) be its foci and P(a cos 0, bsin 0) be any 1 1 , li lit on the cllipse (see Figure 4.3).

Y

P ( a c a 6, bsin 6) b.

FIG. 4.3

4.1 Some Applications of the Conic 121

IPF1l2 = (a cos B + ae)2 + ( b sin B - 0)' = a2(cos2 0 + 2e cos 0 + e2) + b2 sin2 6'

= a2(cos2 6 + 2e cos B + e2) + a2(1 - e2) sin2 B

= a2(cos2 0 + 2e cos 0 + e2 + sin2 6' - e2 sin2 8)

= a2(1 + 2e cos B + e2 cos2 0) = a2(1 + e cos 8)'

Therefore I P F ~ J = ~ ( 1 t e c o s 8 ) .

See also the solution of Example 8 of Section 3.3.3 for an alternative derivation not involving parametric equations.

Similarly ( P F ~ ~ ~ = (a cos B - ae)' + b2 sin" = a2(1 - e cos B)2. Therefore

x2 Y2 The' equation of the tangent to the ellipse - + = 1 at P (a cos 0, bsin 8) is

a2

by equation (26) of Section 3.4.2. Hence distance (Fl TI ( of Fl from the tangent is given by

Ib cos B(-ae) - abl - - ab(1 + e cos 8) IFlTi I =

J(b2 cos2 B + a2 sin2 B) J(b2 cos2 0 + a2 sin2 8)

Therefore

Distance I F2T2J of F2 from the tangent

- - Ib cos 0 (ue) - abl - ab(1 - e cos 6') - (b2 cos2 B + a2 sin2 0) b2 cos2 0 + a2 sin2 8)

Therefore

Hence

Thus light or sound emanating from one focus will be reflected to the other focus. An artificial satellite in orbit round the earth may be used to reflect light from one focus to the other at all times since it will do this at all points of its orbit.


4.1.3 Hyperbola

x2 y2 We recall (Section 3.3.4) that for the hyperbola - - - =

a2 b2 1, IIPFII - (PF2(1 = 2a where FI , F2

are its foci and P is any point on it. This property is used in the location of sources'of sound such as enemy guns. Suppose a transmitter is at an unknown position. Let Fl, F2 be two fixed points. By measuring the difference in time of arrival of a particular sound from the transmitter to the two points, we know 2a (velocity of sound multiplied by the time difference). Hence we know the hyperbola on which the transmitter must lie. Note that we actually know which of the two arms on which the transmitter lies since we know which of the distances IPFlI,IPF2J is shorter. By introducing a third fixed point F3 we now can have one or two more hyperbolas (foci at F l , F3 or F2, F3) and the transmitter lies at the point of intersection of the respective a.rms of all these hyperbolas.

The hyperbola possesses the reflecting property as shown in the next example.

Example 3

x2 y2 , . Show that the tangent at P to the hyperbola - - - = 1 is equally inclined to the lines joining P

u2 b2 " to the foci.

Solution x2

y2 -1~ i th foc i a tF l ( - ae ,0 )andFz(ae ,O) .Le tP (osec8 ,b t an8)be Consider the hyperbola - - - - a2 b2

any point on the hyperbola. See Figure 4.4.

FIG. 4.4

1 ~ ~ 1 1 ~ '= (asec8 + ae)' + b2 tan2 8 = a2(sec2 8 + 2esecO + e2) + a2(e2 - 1) tan2 8

= a2(sec2 8 + 2e sec 8 + e2 + e2 tan2 8 - tan2 8)

= a2(1 + 2e sec 8 + e2 sec2 8) since e2(1 + tan2 8 ) = e2 sec2 0

= a2(1 + e sec 8)2

Thus (PFl I = a l l + e sec 81. Similarly JPF2I2 = (a sec ~ - a e ) ~ + b ' tan2 8 which after simplification reduces to I PF2I2 = a2(e sec 8-

Thus lPF2l = alesec8 - 11.

-4.1 Some Applications of t+,Conic 123 , .

\ I

$!, The equation of the tangent to the hyperbola a t (a sec 0, b tan 0) is

bx - a sin 0 y = ab cos 0 from equation (27) of Section 3.4.2.

Hence distance IPITII of Fl from the tangent

- Ib(-ae) - ab cos 01 - - J j i G G x ) -

Therefore A IFITl) able + cosO1

sinTIPFl = - -

Distance, I F2T21, of F2 from the tangent

- Ib(ae) - ab cos 01

able + cos 01

Jm'

able - cos 01

b(cos01 h - - d7 = sin TI PFi (b2 + a2 sin 0)

Therefore LTIPFl = LT2PF2 and hence the tangent is equally inclined to the lines joining P to the foci.

The above is the reflecting property for the hyperbola. Light travelling from the focus Fi to the point P on the other arm of the hyperbola will be reflected along PL (Figure 4.4). To an observer st L, this reflected light will therefore appear to be coming from F2. Conversely, light coming from L to strike the hyperbola at P will, after reflection, travel to Fl, the second focus, As a result of this, a small hyperbolic mirror is used to solve the practical problem in connection with the parabolic rdecting telescope. (See Figure 4.5).

FIG. 4.5 Combination of parabolic and hyperbolic mirrors in reflecting telescope (hyperbolic mirror ia shown greatly magnified).

124 ~ ~ ~ l i c a t i o n s and General Equations of the Conics

Without the hyperbolic mirror HH an observer would be at F and facing the parabola PP. Thus such an observer would block off a lot of the incident light. When a small hyperbolic mirror is placed so that one of its foci coincides with the focus of the parabola at F , it will reflect all the light incident on it to the other focus F1. F1 can be conveniently positioned behind the parabolic mirror (a small hole at the appropriate place in the wall of the parabolic mirror allows the light through). Thus an observer at Fl can view the entire panaroma.

Exercises 4.1

1 For the ellipse (hyperbola) prove that if the focal chords from any point P on the conic make angles dl,q52 with the axis of the conic, then 2w = $1 + 42 where w is the angle that the normal (respectively tangent) makes with the axis of the conic. Derive this directly without assuming the reflecting property.

2 Find the equations of the bisectors of the angle between the focal chords through any point of P of (i) an ellipse and (ii) a hyperbola.

3 Find, for any conic section, the equations of the bisectors of the angles between the tangent and the normal at any point on the conic.

4 The normal through the point P of the parabola y2 = 4ax meets t l ~ c parabola again at Q. If the tangents at P and Q intersect at R, find the locus of R.

5 In the parabola y2 = 4ax, show that the gradient of the focal chord at any point P is twice that of the tangent at the point.

x2 y2 6 Prove that the product of the distances of the foci of the ellipse -2. + - = 1 from any tangent

a b2 of the ellipse is b2. What is the corresponding result for the hyperbola?

4.2 Polar Equations of a Conic

Let the focus of the conic be at the origin of the x-y co-ordinate axes and the equation of the directrix in the perpendicular form be given by x cos cr + y sin cr = p. Using the distance formula it is easily seen that p is the length of the perpendicular from the origin to the given line whereas cr is the angle this perpendicular makes with O x (see Figure 4.6). Let the point Q with polar co-ordinates (r,8) be any point on the conic.

FIG. 4.6

-

4.2 Polar Equations of a Conic 125

The distance of Q from the directrix = IQKI = lORI - lOSl = p - r cos(8 - a) . But

I Q O ~ = ~ ~ Q K I ' + r=e[p- rcos(8-a) ] .

Therefore r[l + e cos(8 - a)] = ep

Putting B - a = 4 we obtain

where the angle 4 is now measured from the line OR, the perpendicular from the focus to the directrix. OR is thus the (major) axis of the conic. If in Figure 4.6 we had taken Q on the side of the straight line away from the origin, then the distance of Q from the directrix would be given by rcos(8 - a) - p. This would lead to the equation

Thus this latter curve is the same shape as the previous one but centred about 8 = T instead of B = 0.

1 f f (a) r =

1 +ecos$ ' - 5 5 4 S 5 I f 3 f

(b) r = - 5 4 5 - 1 -ecos+ ' 2 2

FIG. 4.7 Sketch of conic r = 1

near (a) 4 = 0 and (b) 4 = ?r 1 f e cos 4

What does the complete sketch look like? We shall take (2) as the standard equation. Since cos(-4) = cos4 the curve is symmetric about the r-axis (as are all of them). Also r -r oo as l + e c o s 4 + 0 . But l + e c o s q 5 = 0 ~ c o s 4 = - ~ .


(i) 0 < e < 1 (ellipse) + a > 1 + 1 + e cos 4 > 0 for all q5 which implies that T is always,finite giving a closed curve which intersects the x-axis at A(&, 0) and B(&, T) in plane polars.

Therefore

FIG. 4.8 Ellipse 1

O < e < l 1 + e c o s + '

(ii) e = 1 (parabola) 1 + cos q5 = 0 6 = n. Therefore

1 FIG. 4.9 Parabola r =

1 + cos+

1 F A = a = - - 1

l + e - 51 since e = 1.

Thus

FIG. 4.10 Hyperbola r = 1

1 + ecosd ' e > l

Therefore

Example 4

Show that the polar equation = a + b cos 0 represents a conic section.

Solution 1 - = a + bcose T + 1 = a r + b r c o s e = a r + b x 1 - bx = a r (1 - b ~ ) ~ = a2r2 = a2(x2 + y2)

i.e. (a2 - b2)x2 + 2bx + a2 y2 = 1.

Case I: la\ # lbl Divide throughout by a2 - b2 and then complete squares

If la1 > lbl this is an ellipse. If la1 < lbl this is a hyperbola since the coefficient of yi is then negative.

Case 11: la1 = jbl (# 0) The equation (i) reduces to

128 Applications and Generai Equations of the Conics

This is a parabola.

0

Exercises 4.2

Identify and sketch each of the following conics. In each case also find its eccentricity.

1 r = 2

2 r = 2

1 + +COSB 1 - 3 COSB

4.3 General Equation of the Second Degree

The most general equation of the second degree is equation (1) of chapter 3 which is repeated here in a slightly different form,

where at least one of the second order terms is non-zero; that is, a , b and c are not all zero. We shall show that such an equation represents one of the conics. It is clear from Chapter 3 that this is the case when b = 0. We may therefore assume that b # 0 and try to put (4) into one of the standard forms of the conics. We shall show that this can be achieved by finding new co-ordinate axes in which the cross product term no longer exists.

4.3.1 The Discriminant

Before looking for new axes, let us see what information we can get from (4). If a = c = 0, b # 0 equation (4) reduces to

z y + 29/22 + 2 f l y + d' = 0

where g' = 9 , f' = $ and d' = $. Thus we have

or equivalently, XY = f w 2

with X = z + 2 f ' , Y = y + 29' and w" Id' - 4 f'g'l. We have already seen in Section 3.3.4 that XY = f w 2 is another form of the equation of the rectangular hyperbola for w # 0. When w = 0 this reduces to the degenrate conic, a pair of straight lines X = 0 or Y = 0.

If either a or c is non-zero, we complete squares. Suppose therefore that without loss of generality a # 0. Divide throughout by u and obtain

4.3 General Equation of the Second Degree 129

which on completion of squares yields

We note that the sign of the coefficient of y2 in (5) would seem to be sufficient to determine completely the type of conic.

Sign of (f - 5) = Sign of (-) = Sign of (4ac - b2)

since 4a2 > 0. Thus the following conclusions appear reasonable:

4ac - b2 > 0 j conic is an ellipse

4ac - b2 < 0 j conic is a hyperbola

4ac - b2 = 0 j conic is a parabola.

These conclusions are in agreement with the already known cases, namely (i) a = c = 0, b f 0 and (ii) b = 0.

However, as much as we are happy to meet once more in another context an old friend in the expression 4ac - b2 = -(b2 - 4 a c ) , these conclusions must remain tentative at present. The reason is that we are in effect replacing x with a new variable, say E l where E = x + $ y . In the case of interest here, namely b # 0, the two variables ( E l y) are not a t right angles and therefore form a set of oblique axes as shown in Figure 4.11.

FIG. 4.11 Oblique axes. OAPB is a parallelogram

It is not clear which conclusions reached using oblique axes will remain valid when rectangular axes are employed. However, we shall retain the discriminant, A E b2 - 4ac, for further investigation.

4.3.2 Rotation of Axes

We prove in this section that it is always possible to rotate the axes of ceordinates so that in the new axes, the general equation of the second degree has no cross-product term.


Let Ozy be the usual rectangular co-ordinate system and suppose the rectangular axes O(q is obtained from Oxy by an anticlockwise~rotation through angle a about the origin (Figure 4.12). Let P be any point on the plane.

FIG. 4.12 Oxy rotated anticlockwise through angle a to obtain O t q

Let its co-ordinates be ( x , y), (<, v), relative to Oxy, Otq axes respectively. Let the distance IOP1 = r and suppose O P makes an angle /3 with the Ot-axis. Drop perpendiculars PQ, P A to Oc,Ox respectively. Then OA = x, P A = y and 0 Q = (, PQ = 7 by definition. From AOAP,

x = r cos(a + /3) = r(cos a cos /3 - sin a sin /3) = cos a ( r cos p) - sin a ( r sin p) = c o s a t - s inaq .

Also y = r sin(a + /3) = r(sin a cos /3 + cos a sin /3)

= s i n a t + cosaq .

Thus

Substituting (6) into equation (4) we have

0 = a(cos a t - sin cu r1)2 + ~ ( C O S a 5 - sin a q)(sin a ( + cos a q)

+ c ( s i n a ( + c o ~ a q ) ~ +2g(cosa t - s i n a q ) + 2f(sina( + cosaq) + d

= (a cos2 a + b sin a cos a + c sin2 a ) t 2 + (-2a sin a cos a + b cos2 a

' - b s i n 2 a + 2 c s i n a c o s a ) ~ ~ + ( a s i n 2 a - b s i n a c o s a + c ~ o s 2 a ) 7 7 2

+ (29 cos a + 2 f sin a )€ + (-2g sin a + 2 f cos a )q + d ~ A ~ ~ + B ( ~ + c ~ ~ + ~ G ~ + ~ F ~ + D = o

where

A = a c o s 2 a + b s i n a c o s a + c s i n 2 a

B = -2a sin cu cos cu + b(cos2 cu - sin2 a ) + 2c sin a cce a!

C = a s i n 2 a - bs inacoea+ccos2a


Now we try to make B equal to zero.

B = -2a sin a cos a + b(cos2 a - sin2 a ) + 2c sin a cos a = 0

+ ( c - a ) s in2a+bcos2a=O (ii)

I fc = a , cos 2a = 0 as b # 0 by assumption (the case b = 0 has already been covered in Chapter 3). Therefore 20 = 5 or a = % radians = 45O. I fc # a , divide (ii) by (c - a) cos 2a(# 0) and obtain

b b - t a n 2 a = --- -. c - a a - c

Hence it is always possible to rotate axes so as to eliminate the cross product term.

Example 5

Find new co-ordinate axes that eliminates the xy term in

(i) x y = 2 (ii) 2x2 + f i x Y + y2 + x - 6 = 0

Solution

If we rotate the axes 'through angle a,

z - + c o s a ( - s i n a q and y + s i n a ( + c o s a q

(i) xy = 2 becomes ( cosa t - s inaq)(s ina t + cosaq) = 2

1.e. cos a sin a t2 + (cos2 a - sin2 a)<q - sin a cos a q2 = 2.

We set cos2 a - sin2 a = 0 to eliminate the cross-product term. That is

t a n a = 1 and a=45O.

Hence putting a = 4 5 O in (iii) yields

(ii) 2x2 + f i x Y + Y2 + x - 6 = 0 This becomes

2(cos cr < - sin a q)' + &(cos a ( -sin a q)(sin a < + cos a q) + (sin a ( + cos a q)2

+ ( c o s a < - s i n a q ) - 6 = O

Multiplying out and collecting like powers of <q yields

(1 + cos2 a + &sin a cos a ) t 2 + (& cos2 a &sin2 a - 2 sin a cos a)(r]

+ (1 + s i n 2 a - &sinacosa)q2 + c o s a t - s i n a q - 6 = 0

(iii)

Weset f i c o s 2 a - f i s i n 2 a - 2 s i n a c o s a = 0 .

132 Applications a n d General Equations of the Conics

Divide throughout by cos2 a ( # 0) wc obtain

d - d t a n 2 a - 2 t a n a ' = 0 = ( I - & t a n a ) ( d + t a n a )

'L'herefore 1

tans=-- or -1.6.

Take 1

a = arctan - = 30° &

'Therefore (iv) becomes

? 7 ! his is an ellipse.

4.3.3 Class i f ica t ion o f general e q u a t i o n of s e c o n d d e g r e e

In this section we shall prove tha t the discriminant does not alter when the co-ordinate axes are rotated and that it is sufficient to help us in determining t,he type of conic.

T h e discriminant of the new equation = B2 - 4AC: where from equation (i) of Section 4.3.2

1 1 1 = -n( l + cos2a) + T b ~ i n 2 a + - c ( l - cos2a )

2 2 2 1 1 1

= --(a + c) - -- ( r : - a) cos 2 a + -b sin 2 a 2 2 2

2 C = a sin' n -- b sin n cos (Y + c C.OS a 1 1 1 - -(1 - cos 2cu) - -h sin 2 a + ;c(l + cos 2a ) 2 2 L 1 1 1

= - ( a + c ) + - ( c - a ) c o s 2 a - -bs in2a 2 2 2

Therefore

(ii)

(iii)

1 8' - 4 A C = [(c - a) sin 2 a + b cos 2aI2 - 4 x -[(a + c) - (c - a ) cos 2 a + bsin 2a]

2

= (c - a)' sin2 2 a + b2 cos"a + 2b(c - a) sin 2 a cos 2 a - {(a + c ) ~

+ ( a + c)(c - a ) cos 2 a - b(a + c) sin 2 a - (a + c)(c - a ) cos 2 a - (c - a)2 cos2 2 a

+ b(c - a ) sin 2 a cos 2 a + b(c + a ) sin 2 a + b(c - a ) sin 2 a cos 2 a - b2 sin2 2 a )

= (c - a)'(sin2 2 a + cos2 2a ) + b2(cos2 2 0 + sin2 2cr) - (a + c ) ~

= (c - a)'+ b2 - ( a + c ) ~

= b2 - 4ac.


Thus the discriminant remains unchanged when the axes are rotated. We offer an alternative derivation of this result which avoids the extensive calculations above.

From (ii) and (iii)

C + A = a + c

C - A = (c -a )cos2a- bsin2a

Hence by (i) and (v)

B' + ( C - A ) ~ = [(c - a ) sin 2a + b cos 2aI2 + [(c - a) cos 2 a - bsin 2a]'

= ( c - uj2(sin92cu + cos2 2,) + 1 r ~ ( c o s 9 ~ a + sin' 2 ~ )

= (c - a j 2 + b2

(iv) '

( 4

'I'tierefore B ~ ~ A c = B ~ + ( c - A ) ~ - ( c + A ) ~

= b2 + (c - a)' - (c + a)' by (vi) and (iv)

= b2 - 4ac.

If we rotate axes to make B = 0, the discriminant A = b2 -4ac = B2 - 4AC reduces to -4AC. a . A > 0 + AC < 0 + hyperbola since one of A, C is positive and the other negative. b. A < 0 + AC > 0 =+ ellipse since both are either positive or negative (and hence can be made

positive). C. A = 0 + AC = 0 + parabola since one only of A, C is zero (both cannot be zero for otherwise

A = B = C = O ) .

Summary

1 Discriminant (B2 - 4AC) 1 AC 1 Conic t v ~ e 1 positive 1 negative 1 hyperbola 1 negative I positive ellipse

1 zero 1 zero I parabola ( TABLE 4. 1 Using the relative signs of A and C to identify the conic

Example G

Determine the type of conic by using the discriminant

(i) z2 + 2y + y2 (ii) x2 + 22y + y2 + 22 - 3y = 0

(iii) z y - y + x = l

Solution

a x 2 + b z y + c y 2 + 2 g z + 2 f y + d = 0

Discriminant A = b2 - 4ac.

(i) a = 1 , b = 1 , c = l , t h e r e f o r e A = 1 2 - 4 ( 1 ) ( 1 ) = 1 - 4 < O ~ e l l i p s e

(ii) a = 1, b = 2, c = 1, therefore A = b2 - 4ac = 22 - 4(1)(1) = 4 - 4 = 0 3 parabola

(iii) a = 0, b = 1, c = 0, therefore A = b2 - 4ac = 1 - 4(0)(0) = 1 > 0 j hyperbola.


Exercises 4.3

In problems 1-6 rotate the ceordinate axe's through an angle a that eliminates the XJ term. Find the equation of the conic in the new ceordinate axes.

1 x 2 + 5 x y + y 2 = 10

2 x2 - , A x y + 2y2 - 4 y = 0

3 11x2 - 1 0 f i x y + y2 + 3 2 d x - 32y + 80 = 0

4 x 2 + 2 x Y + Y 2 + x + y = o

5 x y - y + x = l

6 x 2 - 2 x y + y 2 + 3 x - 2 y + 4 = ~

7 Can equation ( 4 ) represent a circle when b # O? Justify your answer completely.

8 Show that the equation of a circle centred at the origin is not altered by a rotation of axes.

By using the discriminant, determine the type of conic in problems 12-16.

9 2 ~ ~ + ~ y + ~ ~ - 8 ~ - 1 O y - 1 0 = 0

10 5x2 - 2xy.+ y2 + 3x - 2y + 4 = 0

11 x2 - 6 x y + 9y2 + 2x - y = 1

12 By making the change x = (cos a, y = 77 + sin a in equation ( 4 ) show that the discriminant of the resultant equation is (b2 - 4ac) cos2 a . Hence deduce that the oblique axes of Figure 4.11 will also classify the general equation of second degree using the sign of the discriminant.

13 Show that if A = B = C = 0 , then a = b = c = 0 also.

Miscellaneous Exercises 1

1 Show that the polar equation - = a + bsin 0 represents a conic section. Consider the cases r

1.1 # lbl and 1.1 = lbl. Identify and sketch each of the conics in problems 2-5.

2 r = 1

5 + 4sine 3 r ( 1 - sin 0) = 1

By using the discriminant, determine the type of conics in problems 6-10.

6 x 2 + x y + 1 = 0

7 2 x 2 + x Y + y 2 - 1 0 = 0

8 5x2 - 24xy + 12y2 + 3x - 2y + 4 = 0

9 5 x 2 - 2 x y + Y 2 + 3 x - 2 y + 4 = o

10 5x2 + 8 x y - 5y2 + 162 + 20y + 15 = 0

Vectors

5.0 Three dimensional Cartesian Co-ordinate Systems

We are familiar with the two dimensional (plane) Cartesian Co-ordinate system Ox! / . In this system a point in the plane can be uniquely located by an ordered pair of numbers ( x , ? ~ ) called its c* ordinates. In order to describe a point or line in space we need t o extend these ideas lo three dimensions.

Let Ox, Ox, 0 z be three mutually perpendicular lines in space. These are called co-ordinate axes.

FIG. 5.1

The two possible distinct orientations of the axes relative to each other are shown in Figure 5.1. This can be demonstrated by the following proceedure. STEP 1: Stick up the thumb and point it in the direction of the positive z-axis; STEP 2: Point the other fingers of the hand in the direction of the positive x-axis; STEP 3: Confirm that the direction of the positive y-axis coincides with that obtained when the other fingers are now rotated through 90°.

If the right hand is used in the above we obtain Figure 5.l(i) and if the left hand is used we obtain Figure 5.l(ii). Therefore these are called right handed co-ordinate system and left handed co-ordinate system respectively. We shall use only right handed co-ordinate systems throughout in +.his book.

The co-ordinate axes also define three mutually perpendicular planes called co-ordinate planes. These are the - xy plane, perpendicular to the z-axis and containing the x-axis and the y-axis; - x z plane, perpendicular to the y-axis and containing the x-axis and the z-axis; - yz plane, perpendicular to the x-axis and containing the y-axis and the z-axis.

. , ,

136 Vectors , . . . . , .

A point P in space is uniquely determined by an ordered triplet of numbers ( a , b, c) called its coordinates. These are the directed distances from the co-ordinate yz, xz, zy planes respectively to P.

FIG. 5.2

A ceordinate is positive if P is on the same side of the ceordinate plane as is the positive direction of the axis perpendicular to the co-ordinate plane. Thus the real number c, for instance, is positive if P lies on the same side of the xy plane as the positive z-axis. For example if the xy plane is the plane of this page and the positive z-axis points upwards from the page, then c > 0 if P lies above the level of the page but c < 0 if P lies below the level. If P is any point on this paper, then c = 0.

Positive, negative and zero values of a, b are similarly obtained.

Distance of point f rom origin

FIG. 5.3 x

Let P with ceordinates ( x , y, 2) be any point in space. Draw a perpendicular PQ from P to the xy plane to intersect it at Q. Draw perpendiculars from Q to the x- and y-axes to meet these a t R,T respectively. Then OR = x, OT = y and QP = z . By Pythagoras' theorem 10QI2 = 1 0 ~ 1 ~ + I R Q ) ~ = I O R ) ~ + ) o T ( ~ . A ~ S O

Therefore

5.1 Definition and Representation of Vectors 137 8

5.1 Definition and Representation of Vectors

5.1.1 Definition of Vectors

The mathematical modelling of natural phenomena requires the description of various physical quantities. There are two categories of such physical quantities that are of interest to us here; namely, scalars and vectors. Before formally defining these quantities let us consider the following three statements:

a. A town, X , is 60 km from Enugu; b. A town, X , is northwest of Enugu; and c. A town, X , is 60 km northwest of Enugu.

We observe that from the first statement (a) we can only infer that the town, X, lies anywhere on the circumference of a circle with centre at Enugu and radius of 60 km. We therefore do not know its,exact position. We deduce from statement (b) that the town, X , lies on a bearing of 315' from Enugu. Once again, we do not know its exact position.

Statement (c), however, combines (a) and (b) to locate the town X (Nsukka) exactly. We therefore observe that it is necessary to specify both distance (60 km) and direction (northwest) in order to locate the position of Nsukka relative to Enugu.

Such.a physical quantity as the above which requires the specification of both its magnitude and direction for its complete description is called a vector quantity. A physical quantity requiring a specification of only its position on a linear scale in other to give its complete description is called a scalar quantity. The noun 'quantity' is often omitted in these names and we simply call then vectors and scalars respectively.

The special case of a vector with zero magnitude is called a zero or null vector Examples of vectors are velocity, displacement, force, momentum, acceleration Examples of scalars are length, the real numbers, mass, speed, density, price of any commodity,

temperature.

R e m a r k 5.1

The algebra of scalars is the same as the algebra of real numbers (See for example Vol. 1 ) Thus the terms scalars and real numbers will be used interchangeably.

Nota t ion

Recall that the letters a, b , c , . . . , x, y, z are often used to indicate real numbers (scalars). Vectors are_indicated either by using a bold face type a, b, c, . . . or by putting an arrow above the letter 2, b, c', ,. . . or a bar below the letter g , b , c , . . .. A zero vector is denoted as 0, or 6 or 0. The magnitude of the vector a is denoted by la1 or a.

Given a vector x , the vector which has the.same magnitude, 1x1, as x but is opposite in direction to x is denoted by -x.

: Other notations will be defined as the need arises.

5.1.2 Line Segment Representation of Vectors

Given a vector a , this can be represented' pictorially by a line segment P?J whose length IPQJ is equal to the magnitude of a and whose direction P Q is that of a. P is called the initial point and Q the final or end point of the line segment PQ.

138 Vectors

Example 1

Let OABCFEDG be a unit cube (aee ~ i ~ k r e 5.4)

FIG. 5.4

A vector b of magnitude 3 units and which is equally inclined to the co-ordinate axes Oz, Oy, 02 may be represented by the line segment O> of length 3 units in the directio~ OE. The segment P O represents -b.

Note that from the definition of a vector (which specifies only magnitude and direction) any line segment in the direction of the vector and whose length equals the magnitude of the vector represents the~ector. Thus the pictorial representation is not unique. For instance, the line segmente OP, f&Q2, R1 Ra shown in Figure 5.5 each represents a vector in the zy-plane of magnitude 5 unite inclined at an angle of 30° to the positive z-axis.

/ a FIG. 5.5

Note that if we insist that the initial point of a line segment be at the origin 0, then the vzctor ia uniquely represented by o>. More generally, for any point P in space, the line segment O P from the origin 0 to the point P represents a vector whose end point determines the position of P in space. Such a vector is called the position vector of the point P.

5.1 Definition and Representation of Vectors 139

5.1.3 Car tes ian Co-ordinate Representa t ion

The description of vectors and execution of computations involving vectors are made easier or more convenient by the introduction and use of a rectangular co-ordinate system. In such system a vector is represented by an ordered set of numbers called components of the vector. Indeed there are situations when such components have special physical interpretations.

Consider the problem of loading a heavy drum of oil onto a trailer whose flat bed is h metres above the ground. It will require a force equal to the weight of the drum to move it vertically up and load it on the trailer (Figure 5.6(i)). However, by placing a plank inclined a t an angle cr to the ground and resting on the flat bed of the trailer, we now require a force parallel to thc plank to move the drum (Figure 5.6(ii)). This force which is the opposite of the component of the weight parallel to the plank is less than the weight of the drum and decreases as the angle cu decreases. The decrease in ct is achicvcd by using a longer plank.

(ii)

FIG. 5.6

Given a Cartesian co-ordinate system Oxyt , the initial and final points R and S (say) of any line segment representing a vector, v, have unique co-ordinates ( r l , rp, rg) and (s l , s2, s3) respectively (see Figure 5.7).

FIG. 5.7

We may therefork associate with the line segment ~3 and therefore the vector v', the unique triplet (sl - r l , s2 - ~ 2 , s3 - r3) = (pl p21 p3) say, This triplet is also the co-ordinate of the final point P of the position vector which represents v'. (Why is O R S P a parallelogram?).

For convenience, if a vector i7 has associated triplets (?I:, u2, v3) we write C = (vl, v z , us) and use either i7 or (vl , v2, v3) to represent the vector.

Note that in two dimensions we obtain in the above fashion an ordered pair of nuinhers (pl ,p2) say to represent each vector

Equality of vecotrs

Two vecotrs a', 6 with triplets ( a l l a2, as) and (bl, b2, b ~ ) respectively are equal if a1 = h , a2 = b2

140 Vectors

Magnitude of a vector

A vector v' = (vr , vz, v3) has magnitude

Suppose the vector v' = (vl, vz, us) is in a direction making angles al, az, as, say, with the positive directions of the Ox,Oy,Oz axes respectively (see Figure 5.8). LPOA = crl , LPOC = aa, LPOD = C Y ~ . Then

v1 cosal = = V l

Ivl d m 2

These are usually called direction cosines.

FIG. 5.8 J

X

5.2 Algebra of Vectors

5.2.1 Multiplication of a vector by a scalar

The product of a vector v' and a scalar k, denoted by kv', is a vector

where sgn (x), the signum function is defined by

1, f o r x > O

-1, for x < 0

Thus the magnitude of kv', (kv'l is

5.2 Algebra of Vectors 141

and the direction of lev' is t.hat of sgn (k) v' (see Figure 5.9).

I

I I b k -1 0 1

FIG. 5.9

Example 2

The vector 2a has magnitude twice that of a and the same direct i~n as a. On the other hand the vector -4.5b has magnitude 4.5 times that of b and a direction that of -b, that is, a direction opposite t o that of a.

Example 3

In Newton's Second Law we have the equation = ma' which gives the force (a vector) on a particle as the product of mass (scalar) and acceleration (vector).

Remark 5.2

Two vectors a,b are equal, denoted by a = b, if they have the same direction and the same magnitude.

Remark 5.3

Two vectors a, b are said to be parallel if a = kb for some k > 0.

Remark 5.4

A vector whose length is unity is called a unit vector. To determine a unit vector in the direction a (a # 0) we look for a positive real number k such that ka is a unit vector. Thus

Hence the vector ft a or 5 is a unit vector in the direction of a. Note that -h a is also a unit vector but in the direction opposite to a. A unit vector in the direction of a is denoted by a or e',. Thus a vector of length X in the direction of a is Xa or )re',.

In particular vectors parallel to the co-ordinate axes Ox, Oy, 0% will prove useful in subsequent sections.

Notation

The unit vector in the directions Ox, Oy, Oz are denoted by i, j, k respectively.

Thus for example a vector of length A in the direction of i say, is Xi.

142 Vectors

In terms of co-ordinates, if v' = (vl, vz,vs) we want to find the co-ordinates of kc. Suppose these are (wl , w2, w3). For k > 0; the vectors v', kv' are parallel and so have the same direction cosines. Also Iw'l = k 151. Thus

w1 v1 cosff1= --;-= - I w I FI

Similarly it can be shown that w2 = kv2 and w3 = kv3. Thus

kv' = (kvl, kv2, kv3) for k > 0.

For k < 0 let k = - t2 (say). kv' has a direction opposite to that of v' and thus makes angles 180 - cq, 180 - a 2 and 180 - a3 with Ox, Oy, Oz respectively. Since cos(l80 - 0) = - cos0, we have

v 1 w 1 cos al = , and cos(l80 - a l ) = - = - C O S Q ~ lv l I w' I

i.e.

or

Therefore

Similarly for w2 and w3. Thus also kv'= (kvl, kv2, kv3) for k < 0. For k = 0, the same equation is satisfied. Hence

kt7 = (kvl, kvz, kvs) for all k E R.

One way of defining the sum a'+ of two vectors a', g in that order is to do this wing the pidorial representation 18% Figure 5.10).

(9 FIG. 5.10

For the vector g we take line segment representing 6 which has its initial point at the end of the line segment for a'. The sum a' + 5 is represented by the lin5 segment joining the $itd ~ o i n t of the line segment for a' t o the final point of the line segment for b. Thus the vectors a', b, a'+ b form the three sides of a triangle. This result is called the triangle law of vector addition.


R e m a r k 5.5

The expression of the form 'the line segmefit A% representing the vector a" appears frequently. For brevity, we shall where convenient shorten this to 'the vector A%'.

Prover t ies of vector addit ion a+o '=a a'+ (4) = o' a'+b'=b'+a' commutative law (a'+ ;) + c'= a'+ (;+ EJ associative law k(a' + ;) = ka' + kg for all k E R

PROOF. ( 1 ) From the triangle law if i = 6 then the end point of 6 coincides with its initial point and_ the triangle collapses to the vector a'. Thus a'+ 8 = a'. (2) If b = (-a') then $ and a' have the same magnitude but opposite directions. Thus the end point of ; coincides with the initial point of a'. So that a'+ ( - i f ) = 6. ( 3 )

FIG. 5.11

In Figure 5.11, PQ represents the vector ii, PS the vector ; and PQRS is a parallelogram. Thus - QR = ; and % = a' also. Hence

Also - - - P R = P S + S R = ; + Z .

Therefore a ' + g = ; + z = P R .

R e m a r k 5.6

Figure 5.11 shdws that the sum$+ a of two vectors i f , g is represented by the diagonal of a parallelogram formed by the vectors a', b and with a', ;, a'+ ; passing through the same point. This is called the pamllelogram law of vector addition.

( 4 ) Vector addition is defined only for pairs of vectors. Thus to add three or more vectors we can only add them in pairs. The expressions (a'+;) +Z, a'+ (;+q show two such ways of adding a', g, ?in pairs. The essence of the result is that the sum of the three vectors is independent of the sequence of these pairs. Thus

144 Vectors

FIG. 5.12

From Figure 5.12 - A D = A C + C D = ( Z Z + B C ) + C D

= ( a ' + Q + c :

Also

Hence ( a + S ) + c'= a'+ ( 6 + q .

By using property (3)

a + ( S + q = ( 6 + q + a = 6 + (?+a) by (i).

Other forms can be obtained similarly. ( 5 ) To prove that multiplication by scalar is distributive over vector addition, namely

we refer to Figure 5.13 which illustrates the sums a'+ 6 and ka'+ k6 for k > 0.

A

FIG. 5.13


Let = a', BC = and P& = ka', &R = kc. Thus AZ = a' + g and = ka' + kg. Since - AB [ I PQ and BC 11 m, then angle ABC = angle PQR. Also (WI = k \ m l and I&RI = k I E I . Hence AABC is similar to APQR. Therefore In] = kJACI and CTB = R ~ Q SO that PR 1 1 E. Hence = k x i.e.

ka'+ kg= k(a'+ 6). The statement is trivially true for k = 0. The case k < 0 is left as an exercise.

5.2.2 Addition of Vectors

Subtraction of vectors

~ e f i n e z - ; := a'+(-;) Thus to subtract ;from a', we change the sign of $ and add the result to a'.

Vector equation of a straight line

Find the equation of a line through the point Po and parallel to the vector ii.

Solution

Take a representation of a' which passes through the point Po. Let 0 be the origin of co-ordinates and P any point on the required line. Let the position vectors of Po, P be Fo, F respectively (see Figure 5.14).

FIG. 5.14

Now P;P 1 1 a' therefore P ~ P = ta'. That is 7- Fo = ta' or

q t ) = F , + ta.

The notation ?(t) being used to emphasize that the position vector of P depends on the value of t. If t = 0, F(0) = To and P coincides with Po. If t > 0, P;P is in the direction a' and P lies to the right of Po in the diagram. If t < 0, P;P is in the direction of -a' and P lies to the left of Po in the diagram. Thus the entire line is generated for t E ( - o o , ~ )

146 Vectors

Example 4

Internal division of a line segment Let a', b be positioil vectors of the points A, B respectively in space relative to any origin 0. Determine the position vector of the point P which divides the line joining A, B internally in the ration of rn: n. Solut ion

Let the position vector of P relative to 0 be F. Now A> I( ~3 therefore A> = k ~ % for some k > 0 IA>J nz * k=-- - . - (P>(

Therefore m -

A>= -PB or n ~ > = m ~ % . n

~ u t ~ > = ~ - a ' a n d ~ % = i - r ' Therefore

n(?-a')=rn(i-r") or ( n+m)?=na '+mC

0

If n = m, P becomes the midpoint of AB and

Resul t ( ~ i c l ~ o i n t theorem)

The line joining the midpoints of two sides of a triangle is parallel to the third side and equal to half its length

PROOF. Let A, B,C be any triangle in space. Let a, b, c be the position vectors of A, B,C respectively relative to some origin of co-ordinates. If D, E are the midpoints of AB, AC respectively,

a then their position vectors are


Therefore

Hence

We now turn our attention to deriving the results of addition of two vectors when these are expressed in ceordinate form. Thus given two vectors a = ( a l , a2, as) , b = ( b l , b 2 , b3) say, we wish to derive their sum a + b in co-ordinate form. To do this we first obtain an expression for v = ( v l , v ~ , v3) as a sum of vectors involving the unit vectors i , j, k . In Figure 5.15 O> is the position vector of v, lOAl = v l , (081 = v2, lOCl = 213. But 02 is a vector in the direction of i and of length v1

FIG. 5.15

Thus 02 = v l i . Similarly 0% = v l j and 02 = v lk . Now

Hence V = (v1,v2,v3) = v l i + v j + v g k

If a vector is written in the form (i), v l , vz, 213 are called its components in the direction i , j, k respectively.

148 Vectors

By (ii)

Thus

or

by associativity and comutativity

( a ~ , az, as) + (bl, bz, b3) = (a1 + bl, a2 + bz, a3 + b3).

Example 5

If a = 3i + 2j + 4k and b = -4i + 2j + 5k find (i) a + 2b (ii) 3a - b

Solution

a + 2 b = 3 i + 2 j + 4 k + 2 ( - 4 i + 2 j + 5 k ) = 3i + 2j + 4k - 8i + 4j + 10k by property (5) = (3 - 8)i + (2 + 4)j + (4 + 10)k

= -5i + 6j + 14k

Example 6

If a = -3i + 4 j + 5k, b = 2i + 3j - 5k and c = -2i + 14j show that c is parallel to a + b

Solution

a + b = ( - 3 i + 4 j + 5 k ) + ( 2 i + 3 j - 5 k ) = (-3+ 2)i+ ( 4 + 3 ) j + (5 - 5)k

= - i + 7 j

But c = -2i + 14j = 2(-i + 7j) = 2(a + b).

Therefore c ( 1 ( a + b) since c = k(a + b) with k > 0.

Example 7

Let A(-3, -7,9) and B(7,8, -6) be points in space. If C, D divide AB in the ratio 2:3 and 3:2 respectively, show that AZ = D%.


Solution

Let a, b be the position vectors of A, B respectively. Then ~ l l e position vector r of any point P dividing A B in the ratio m: n is given by

For the point C, rn = 2, n = 3 and the position vector of c of C is

For the point D, m = 3, n = 2 and the position vector d of D is

Thus

Therefore ~3 = ~ 3 .

Example 8

Find the point of intersection of the line r(A) = (4 ,3 ,1) + A(3,1,2) with the line through the points with co-ordinates (2, -1,O) and (4, -7,2). Solution

For the equation of the line through (2, -1,O) and (4, -7,2): direction a = (4,-7 ,2) - (2,-1,O) = (2,-6,2).

150 Vectors

Therefore equation of line is r(t) =,(2, -1,O) + t(2, -6,2)

Equation of given line is r(A) = ( 4 , 3 , l ) + X(3,l, 2) (ii)

At point of intersection, if any, r(t*) = r(X*)

where 1" , A* are the values of 1 , A respectively which give the co-ordinates of the point of intersection.

Therefore (2,-1,O) +te(2,-6,2) = (4,3,1) -I- X*(3,1,2)

or ( 2 + 2 t * , - 1 - 6 t * , 2 * ) = ( 4 + 3 X * , 3 + X * , 1 - k 2 ~ * )

(iii)

(iv)

(v) -

W must solve equations (iii) - (v) simultaneously for t*, A * . Let us use (iii) and (v) t o solve and h e n check if these values also satisfy (iv).

(iii) and (v) 2 = 3 + X * + X * = - 1

Substituting this into (v) we have

1 21" = 1 + 2(-1) = -1 + t* = --

2 '

Check in equation (iv):

LHS = -1 - 6( - i ) = 2 and RHS = 3 + (-1) = 2. Hence LHS = RHS and equation (iv) is satisfied.

Therefore the point of intersection is r(t* = -:) = r(X* = -1).

Exercises 5.2.2

1 By completing the parallelogram OADB, prove that o ~ + o > = 2 0 2 where C is the midpoint of AB.

2 A, B , C and 0 are any points in space. Prove that 02 -k 0% = 20%' + ~2 -k c>.

3 Let D, E be points on the sides AB, AC respectively of A A B C such that D divides AB in the ratio m: n and E divides A C in the same ratio. Prove that DE is parallel to B C and of

m length -)BZ).

m + n


Obtain the equation of the line through the point A(l, -3,2) and parallel t o the vector (-12, 4,3)

(i) Yerify that the point (-23,5,8) lies on this line. (ii) Find the co-ordinates of the point on the line with x-co-ordinate 3.

(iii) Find the co-ordinates of two points on the line each distant 5 units from A.

Prove the distributive law k(a + b) = ka + kb for the case k < 0.

Given that town B is 150 km North of town A and town C is 100 km southwest of town B , determine the position of C relative to A. A boy can swim with a speed of 26 metres/minute in still water. He wants to swim across a 150 m wide river from a point A to a point B directly opposite on the other side of the river. The river flows at 10 metres/minute.

a. If he always swims in the direction parallel to AB, find how far he lands downstream of B.

b. In what direction relative to the bank of the river must the boy swim so as to cross directly from A to B?

Let A, B, C be points with position vectors a, b, c respectively. Show that A, B , C are collinear if and only if there are non-zero constants a, p , 7 such that a a + p b + y c = 0 with a + @ + y = 0.

5.2.3 Scalar p roduc t of two vectors

We first introduce a concept which will be used in defining the products (both scalar and vector products) of two vectors.

Angle be tween two vectors Given two vectors, we select a pictorial representation of both in which they have a common initial point. With coincident initial points the two vectors/line segments now lie in a plane since any three points define a plane: in this case the three points are the coincident initial point, and the two final points of the vectors.

FIG. 5.16

The angle, 8, between the two vectors is defined as the smaller of the two angles between the directions of the vectors (see Figure 5.16). The line segment, dotted in Figure 5.16, which joins the final points of a and b represents the vector b - a. By the cosine law

i.e

(bl - a ~ ) ~ + (b2 - a2)2 + (bs - as)2 = a: +a: + a; + b: + b i + b i - 21al Ibl cos 8,

152 Vectors

Hence

Therefore

We can now define the scalar product of two vectors a and b .

DEFINITION. Let a, b be any two vectors, the scalar product or do t p roduc t of a and b in that order, denoted by a . b is given by

'where 9 is the angle between a and b .

We note from the definition that (i) a - b = 0 if at least one of lal, lbl, cos9 is zero. Thus if either a = 0 or b = 0 then a . b = 0. Also if 0 = $, cos = 0 and a . b = 0 even when la\ # 0 and Ibl # 0. Such vectors, that is with the property a . b = 0, are said to be perpendicular or orthogonal. In particular

(ii) If two vectors are parallel, the angle between them, 0 is zero and cos 0 = 1. Thus in particular a a = la1 la1 - 1 which implies

Also

(iii) a . b = JaJ Jbl cos 0 = la1 lbl . atbt + a 2 b ~ + a3b3

lal Ibl = a l b l + azb2 + asb3

Thus

Proper t ies of scalar p roduc t 1 a . b = b . a 2 a . ( k b ) = ( k a ) - b = k a - b f o r k € R 3 a . ( b + c) = a . b + a . c distributive law

PROOF. Properties (1) and (2) for k 2 0 follow directly from definition. For k = -w2 < 0 say,

l a . (kb)l = la1 IkbJ cos(l80 - 0) = la1 lkl Ibl(- cos 0)

5.2 Algebra o f Vectors 153

where 0 is the angle between a and b. Therefore

la. (kb)l = /alw2/b/(- cos 8) = -w21al /b( cos 0 = klal lbl cos 0 = ka-b .

0

The distributive law can be extended to the following:

Componen t s of a vector We now generalize the idea of the component of a vector defined earlier in the directions i, j , k to the component of a vector in an arbitrary direction.

The vector component or vector projection of a vector a in the direction of the unit vector il is

( a . G)&

The scalar a . u is called the scalar component or scalar projection of a onto G . Note that the above definition is consistent with the earlier one since if a = a l i + a j + a3k then

Similarly a . j = a2 and a. k = a3 as before.

Example 9

Compute the scalar product of a = 2i - 3j + 5k and b = -6i + 9j - 15k. Also determine the angle between them.

Solut ion a. b = (2i - 3j + 5k) . (-6i + 9j - 15k)

= 2(-6) + (-3)(9) + 5(-15) = -114

Therefore

154 Vectors

Hence e = COS-'(-~) = T.

0 = ?r + a, b are in opposite directions which is correct since b = -3a!

0

Example 10

If the cosine of the angle between u = i + 2j + 2k and v = i - 4 j + pk is 5 , find the value of the parameter p.

Solution

Let the angle between u and v be 8.

thus

Squaring we have

Therefore

We check these values in (i) since the squaring can introduce spurious 'solution'

4 2 ($) - 7 p = - * LHS of (i) =

3 = -1 Not a solution

2(8) - 7 p = 8 * LHS of (i) =

d m = 1 Solution

Therefore p = 8.

Vector equation of a plane

DEFINITION. Given a non-zero vector n, a plane in R3 is the set of points, S, such that the line segment/vector joining any two points in S is perpendicular to n. Thus for a, b E S, n . (a - b) = 0.


See Figure 5.17. The vector n is called the nonnal to the plane.

FIG. 5.17

Remark 5.7

Fkom the definition 1 -n is also a normal to the plane; 2 n a = n . b for all a, b E S. Hence S consists of all points the scalar projections of whose position vectors in the direction of n are all equal (and equal to OT in Figure 5.18).

FIG. 5.18

To obtain the equation satisfied by the position vectors r of all pointe in the plane coneider a point ro and the plane through ro = (xo, yo, zo) with normal n = (A, B, C). Let P with position vector r = (x, y, z ) be any other point in the plane. Then P?O Po n and thus

n . (r -ro) = 0 =$ n - r = n.ro = const.

In terms of components we have

r AX + Bv + Cz = constant 1

Example 11

Find the equation of a plane through the point (1, -2,3) and with normal 4i + 3j + k.

Solution

Fixed point on plane, ro = (1, -2,3). Normal to plane n = (4,3,1). Let r = (x, y, z) be an arbitrary point in the plane. Then

156 Vectors

5.2.4 Vector product of two vectors

DEFINITION. The vector product or cross product of two vectors u, v in that order, denoted by u x v o r u A v , is

U A V = lulIvIsint9A

where 0 is the angle between u, v with u, v, A forming a right handed system, i.e. a rotation of the hngers of the right hand frbm u to v leaves the thumb pointing in the direction of A. Thus the milgnitude of u A v is

(u A vl = I u I I v I sin 0.

Remark 5.8

From the definition

(i) u A v = 0 if either lull I v I or sine is zero. Thus if either u = 0 or v = 0, then u A v = 0. Also if 0 = 0 or n, sine = 0 and u A v = 0. Hence a A a = a A (-a) = 0. In particular i A i = j A j = k A k = O .

(ii) i Aj is a vector in the direction of k and so i A j = lil Ijl sin $ k = k. Similarly j A k = i and k A i = j .

(iii) u . (u A v) = lul Ivl sin 0 u . ?t = 0 and v . (u A v) = lul Ivl sin 6' v . ?t = 0 since ?t is perpendicular to both u and v.

(iv) I U A V I is the area of the parallelogram formed with u, and v as adjacent sides.

Properties of vector product 1 b A a = - a A b . 2 a Ab = (azb3 - a3b2,a3bl - alb3,alb2 - a2bl). 3 (Xa) A b = a A (Xb) = X(a A b) for any real number A. 4 a A ( b + c ) = a A b + a A c distributive law.

PROOF. 1 Ib Aal = laA bl but as the roles of a, b are interchanged in the right handed system, b A a points in the direction' opposite to a A b.

2 Let a A b = (x, y, z). We know from Remark 5.8 (iii) that a (a A b) = 0 and b . (a A b) = 0 Hence alx + azy + a3z = 0 and blx + bzy + b3z = 0. If a, b are both non-zero and non-parallel we can solve these equations (see Example 14) to obtain


in terms of z where alb2 - a2bl # 0. From the definitions of a . b and a A b, since cos2 0 +sin2 0 = 1, we have

Thus la bI2 = Ja)21b(2 - ( a . b)'

i.e z2 + y2 + z2 = (a: + a; + ag)(b: + b; + bz) - ( a ~ b l + azbz + ~ 3 b 3 ) ~ = (azb3 - a3b2)2 + (a3bl - alb3)2 + (alb2 -

Substituting above for x and y yields

Thus a A b = f ( ~ b 3 - a3b2, a3bl - alb3, alb2 - a2bl).

To determine which sign to use, we substitute the special case a = i = (1,O, 0)) b = j = (0,1,0) so that a A b = k = (0,0,1). This gives (0,0,1) = f (O,O, 1). Hence the correct sign is the positive sign. Therefore

a A b = (a2b3 - a3b2,a3bl - alb3,albe - a2b1).

Now we check if this equation is also correct for the simple cases we had left out of consideration. For the case of one or both of a , b being zero, then clearly a A b = 0 by definition and a A b = (0,0,0) agrees with (*). If a ( 1 b then b = Aa and a A b = 0 since 0 = 0. This is also consistent with (*) since b '= (Aal, Aa2, Xu3). Hence in all cases

3 (Aa)Ab=aA(Ab) = A(aAb) for all A .

Similarly it can be shown that a A (Ab) = A(a A b).

4 a A ( b + c ) = a ~ b + a ~ c As an exercise, use components to derive this result.

Remark 5.9

(i) Particular instances of property 1 are

(ii) The distributive law can as usual be extended to

158 Vectors

Triple p roduc t s r Since a A b is a vector, its product (both s'calar and vector) can be taken with another vector c, aay to obtain ( a A b) A c called vector triple product or (a A b) c called scalar triple product.

Example 12

G i v e n u = 3 i + 4 j - 7 k , v = - i + 2 j + k a n d w = i + 2 k 1 f i n d

(i) u A v (ii) ( u A v ) w (iii) (u A v) A w (iv) u A (v A w)

Solut ion

GI u A v = (3,4, -7) A (-1,2,1)

= - (-7)(2), (-7)(-l) - (3)(1), 3(2) - 4(-1))

= (l8,4,lO) or 18i + 4j + 10k

(ii) ( u A v ) w = (l8,4,lO) - (1,0,2) = 18(1) + 4(O) + lO(2) = 38.

(iii) (u A v ) A w = (18,4,10) A (1, 0,2)

= (8 - 0,lO - 36,O - 4) = (8, -26, -4)

6.) V A W = ( - 1 , 2 , 1 ) ~ ( 1 , 0 , 2 ) = ( 4 - 0 , 1 + 2 , 0 - 2 ) = ( 4 , 3 , - 2 )

Therefore u A (v A w) = (3,4, -7) A (4,3, -2) = (-8 + 21, -28 + 6,9 - 16) = (13, -22, -7) # (U A V) A w

Example 13

Find the equation of the plane passing through the three points A(1,2- I), B(-3,4,0) and C(2,1,1).

Solut ion

The vectors A%, A*, BZ are co-planar. The vector product of any two of these will give a normal n to the plank since the vector product of any two vectors is perpendicular to the two vectors and thus to the plane containing them. Let a, b, c be the position vectors of A, B , C respectively. ,

Therefore , n = A 3 A A* = (-4,2,l) A (1, -1,2)

= (5,9,2).

Hence equation of plane through a and with n as normal is


Example 14

Use vector methods to solve a lx+a2y+a3z = 0 blx + bzy + b3t. = 0.

Solut ion

The given simultaneous equations may be written in the form

a . x = 0 b . x = O

where a = (al, aa, as), b = (bl, bz, b3) and x = (x, y, z). These latter equations imply x l a and x l b . Therefore

x = X(a A b)

is a solution for any real number A.

0

Exercises 5.2.4

I f a = - i + 2 j + k l b = 3 i + 4 j + k , c = i - 4 j + 3 k f i n d (i) a ~ b (ii) ~ A C .

Show that the vectors p = (3, -2, O), q = (2, -4, -2) and r = (1,2,2) are coplanar.

Determine which sets of vectors are coplanar (i) (0,2,-3), ( l , l , - l ) , (-1,9,0)

(ii) (1,213)~ (0, 3, -l)i (2i l ) (iii) (-2, 1, -11, (1, 2,3), (-1,3) 2)

I f u = 4 i - 3 j + 5 k , v = i + 2 j + 5 k a n d w = 7 i - 2 j f i n d (i) u . v

(ii) u . (2v + 3w - 5u) (iii) ( u + 2v) . (w - 2v).

Find the equation of the plane through the points (1,2,3), (0,3, -1) and (2,1,1).

Find the equation of the plane through the points (O11, 2), (1,0,1) and (2,1,0).

Find the equation of the plane through the points (3, -1,2), (4,0,2) and (1,1,1).

Show that r(X, p) = a + Xb + pc where X I p are real numbers is a plane containing b and c.

Solve

10 Solve

11 a. P r o v e a ~ ( b A c ) = ( a A b ) ~ c f o r a l l a , b , c . b. Prove that if the vectors a, b, c are coplanar, then a ( b A c) = 0.

Vectors

Show that a = 2i - 4j + k and b = 3i - j - 10k are orthogonal.

Find the angle between (i) a = i + 2 j ' - 2 k a n d b = - 2 i + 3 j + 6 k (ii) u = 2i + 6j - 3k and v = -7i + 4j + 4k a. If p is the vector projection of the vector u = 3i - j + 2k onto the vector v = 4i + 2j - 3k

(i) find p

(ii) show that u - p is orthogonal t o v

b. Show that the vector projection of a in the direction of b is

Consider the line L with equation r(t) = a+ tu . To find the distance from the point c to the line, use the methods of problem 14a. to find the component of c - a perpendicular to the line. Illustrate your methods for the case a = 2i + j + 3k, u = i + j - 2k and c = -i + 3j + 2k. What is the foot of the perpendicular from c to the line L? Repeat problem 15 using an arbitrary point r(t0) on the line in place of a to show that the results are independent of the choice of point r(to) on the line.

Calculus of vector functions of a real variable

In real life some vectors are functions of one or more real variables. For example, the velocity of a stone thrown vertically upwards depends on, i.e. changes with, time. Also the velocity of a stream varies with both time and position. However, in this section we shall consider only vector functions that depend on one real variable only.

5.3.1 Differentiation of vector funct ions of a real variable

Let u(t) denote a vector function of one real variable t . For any arbitrary number to , we define

lim u(t) = lim (ul(t), uz(t), u3(t)) i - t o t - l o

: = ( lim u,(t), lim u2(t), lim u3(t) t - t o t - t o i - t o

A function u(t) is continvovs at t = to if

that is

Thus by equality of two vectors

Hence, a vector function is continuous at to if and only if each of its components is continuous a t to. Also as for functions of a real variable a vector function is said to be continuous in an open interval if it is continuous a t every point in the interval.

Example 15

If u(t) = (t2 + 1, ei , sin 3nt)

5.3 Calculus of vector functions of a real variable 161

(i) find limt+2 u(t) (ii) show that u(t) is continuous everywhere.

Solut ion

lim u(t) = (lim(t2 + I), lim et , lim(sin 3 d ) ) 1-2 t+2 2 t-2

= (5, e2, 0).

(ii) The components of u(t), namely, t2 + 1, et,sin 37rt, are continuous everywhere. Hence u(t) is continuous everywhere.

Let u(t) be a continuous vector function of the real variable t and u(to), u(to+At) be the values of u at two neighbouring points to and to + At . Suppose these vectors are represented pictorially by the line segments A% and as in Figure 5.19.

FIG. 5.19

Let u(to + At) - u(to) be denoted by A u which is therefore represented by B?. Thus A u = u(to + At) - u(to) and

1 -(Au) = u(to + At) - u(to) At At

As t varies, the final point C of u(t) traces out a curve or path. If At --, 0, A u --, 0. A u

If the lim - exists, the limit is called the derivative of the vector function u(t) at t i.e. A t 4 0 At

du u(t + At) - u(t) - = lim dt at-o At

du du where - denotes derivative of u with respect to t . - is also denoted by ul(t). Note that the ratio

dt dt A u 1 - = -(Au) is a vector (parallel to the line segment B? in Figure 5.19) and hence its limit is At At also a vector. This limit vector is tangent at B to the path traced out by C.

From the description of limit, it follows that

dul duz du3 dt dt ' dt ' dt dt dt dt

- i+-j+-k.

Thus, for example, for u = (t2 + 1 ,e t , s i n3d ) ,

du - = (2t,et,37rcos3nt). dt

162 Vectors

Properties o f the derivative , .

For any k E R, a any constant vector, u( t ) , v ( t ) are differentiable vector functions and d( t ) any differentiable scalar function,

(4) d z(a) = 0

(5) d du -(ku(t)) = kx dt

(6) d du dv - ( u + v ) = - + - dt dt dt

(7) d d 4 du -(du) = -u+d- dt dt dt

(8) d du dv - ( u . v ) = - - v + u . - dt dt dt

(9) d du dv - ( u A v ) = - A V + U A - dt dt dt '

5.3.2 Integration o f vector function o f a real variable

DEFINITION. Let u( t ) be a continuous vector function of the real variable t , define the integral or antiderivative of u( t ) by

Rom this definition and the properties of integrals of scalar functions we can deduce the following properties:

In what

/ Odt = a where a is a constant

follows and in the exercises we assume that the position vector r ( t ) , velocity v ( t ) an' acceleration a(t) are related by

d d v ) = ( t ) , a(t) = -v(t).

dt

These concepts are discussed more fully in the next chapter.

Example 16

The velocity, v ( t ) of a particle at time t is given by

v ( t ) = 2 i + cost j - sint k

If at time t = 0, the particle is a t the origin, find it8 position vector at t = 5.

5.3 Calculus of vector functions of a real variable 163

Solution dr

v(t) = - = 2 i + c o s t j - s i n t k dt

Integrating with respect to t we obtain

r(t) = 2t i + sint j + cost k + c

where c is a constant vector (integrating constant).

But at t = 0, r = 0 Therefore

Therefore

which implies c = -k. Therefore r(t) = 2t i + sint j + (cost - 1) k. A t t = $ ,

lr lr lr lr r(?) = 2(?) i + sin - j + (cos - - 1) k

2 2

Exercises 5.3

Given that 4 = et , u(t) = sint i + cost j + 3 k and v(t) = t i - 2 k. Find

d -(u A v). dt The velocity of a particle at time t is v(t) = t i + t2 k. If at time t = 0, the particle is at the origin, find its position vector. The position vector r(t) of a particle is given by r(t) = 2 cost i + 2 sin t j + t k, find its velocity, speed, and acceleration. The acceleration of a particle is a(t) 2 -10 k. At time t = 0 the particle is at the origin and has velocity 100 k. Find its displacement as a function of t and its value when the velocity is 0. For each case'below, find i and verify that r - r = 0. a. r = a(cos 8, sin 8) b. r = a ( c o s e c o s ~ , c o s ~ s i n ~ , s i n 8 ) . Check that lrI2 = a2 in each case.

Miscel laneous Exercises

1 Obtain the vector equation of the line through the point A(-1,0,3) and parallel to the vector u = (5,3, -4). Find the point where this line intersects the xy-plane.

2 Obtain the equation of the plane through the point (1, -1,2) and normal to 2 i - 3 j + 4 k.

3 Use component form of vectors to prove that a A (b + c) = a A b + a A c .

164 Vectors

G i v e n t h a t u = 3 i - 2 j + 4 k 1 v = - i + k , w = - 2 i - j + 3 k 1 f i n d (i) u . w (ii) ( u A v ) . w ' (iii) ( u A v ) A w .

Prove that for any vectors u and v, ( u A v) . u = 0.

Compute the scalar and vector projections of u onto v and the angle between u and v if a. u = - i + 3 j , v = i - 2 j + 2 k , b. u = 1 2 i + 3 j - 4 k , v = - 3 i + 4 k . Find the equation of the plane through the points (i) (2,0931, (191, -1) and (-21 1,2), (ii) (3,1,2), ( - 1 ,2 , l ) and ( l , - 1 ,4 ) . Find the point of intersection of the line r(t) = (1, -2,O) + t (2,1, -1) and the plane through the point (3, - I l l ) with normal n = (1, -1,O).

d Given that 4 = s i n t , u = t i - j and v = c o s t i + j + s i n t k f i n d - ( ~ u A v ) .

dt The position vector r of a particle is given by r( t ) = t i + 2t2 j + k , find the velocity and acceleration of the particle. The position vector r(t) of a particle is given by r = (3sin 2t,3cos 2t, 2t), find its velocity v and acceleration a. Show that the acceleration is orthogonal to the velocity. The velocity v(t) of a particle is given by v(t) = (ucos8, usine - gt, 0) where u,8 and g are constants. Find the displacement vector of the particle if it is initally at the origin. Show that the path of the particle is a parabola. The acceleration of a particle initially at the origin with velocity 2j is cost i - 2 sin t j, find the position of the particle as a function of time. Identify the path of the particle.

Kinematics of a Particle I n this and the next chapter we shall study the motions of solid objects and bodies. To do this we build a mathematical model of such motions, We then solve the equations arising from this model. How good the model is is determined by how well the results mimic reality. In this chapter we discuss the motions of objects called particles without consideration of the forces producing such motions. Such a study is called kinemalics of a particle.

6.1 Basic Concepts

Concept of a part ic le Consider two friends Ada and Dupe who are standing at a bus stop on the side of a straight road. A bus arrives and Dupe boards it. As it drives off the two friends wave to each other.

As observed by Ada, the bus loomed large when it was at the bus stop: she could identify the passengers and their individual motions including the waving hand of Dupe. However, as the bus drove farther and farther away from her, she could no longer obseerve the individual passengers and their movements; moreover the bus appears smaller and smaller until it becomes a dot a t the horizon.

Let us contrast Ada's view of the bus with that of Dupe. For the latter, the bus appears of constant size and she continues to observe the passengers and their individual movements as clearly as Ada had seen them while the bus was standing at the bus stop.

We note that in reality the motion of the bus and its passengers is independent of the two observations. However, what each of the friends actually observes is dependent on their respective distances from the bus and its passengers. Important therefore to each view is the size of the observed object in comparison t o its distance from the observer. If an object is such that its size is small relative to the distances involved in a description of its motion by an observer, then for such a description of its motion by an observer the object is called a particle and its position is represented in our mathematical model by a point. The position of the observer is called the reference or observation point.

The following are cases where the moving objects can be regarded as particles: motion of particle reference point

(i) earth round the sun centre of the sun (ii) pendulum (i.e., bulb tied to end of a long string) fixed end of string (iii) stone thrown into air point on the ground (iv) aeroplane flying from Enugu to Lagos Enugu airport

F r ame of references A co-ordinate system with origin at the point of observation with respect to which the movement of an object is described is called a l m m e of reference. We shall use only right-handed Cartesian co-ordinate systems.

Displacement Vector Given a rectangular co-ordinate system with origin 0 consider a particle at a point P in space a t time l o with position O> = r(to). If the particle moves to another position Q at time t > t 0 , then

166 Kinematics of a Particle 1'' *I '

its new position vector is OQ = r( t l ) . The vector = r(t1) - =(to) is called the dispJa'cemeni vector (or displacement for short) of the particle as it moves from P to Q (see Figure 6.1). Thus starting from time to , the position vector of a particle at time t is r(t) and its displacement vector is r(t) - r(to).

FIG. 6.1

Note that as t varies the end point of the vector r(t) traces out a particle path or trajectory.

Example 1

Let the position vector of a particle P be given at time t seconds by

Find (i) the position vector of P at t = 2 (ii) the displacement of P at the end of 2 seconds.

Solution

(i) When t = 2, r(2) = 2(2) i + (2' - 1) j + 3 k = 4 i + 3 j + 3 k.

(ii) For t = 0,

Therefore displacement a t the end of the 2nd second ig given by

Example 2

The path of a particle is given by

r(t) = ( t 2 - t ) i + (t2 + t ) j

a. When is the particle a t (0,2,0)?

b. Does the particle return to its position a t t = 0 for any positive time?

6.1 Basic Concepts 167

.- . a. Hence the particle is a t (0,2,0) when

and t 2 + t = 0

the r component being zero at all times. (i) =j t = 0 or 1 (ii) * t 2 + t - 2 = ( t + 2 ) ( t - l ) = O * t = - 2 o r l . Hence (i), (ii) are satisfied simultaneously for t = 1. Thus the particle, is a t (0,2,0) when t = 1. b. When t = 0, r(0) = (0 - 0) i + (0 + 0) j = O The particle returns to the origin again when

z = t 2 - t = 0 (iii)

and g = t 2 + t = 0 ( 4

(iii) a t = 0 or 1 and (iv) *t = 0 or - 1. Thus (iii) and (iv) are satisfied simultaneously only for t = 0 (initial time). Hence, the particle d m not return to. the origin.

Velocity a n d acceleration vectors

DEFINITION. The velocity, v(t) of a particle is the rate of change of its &placement with respect to time. Thus

Hence, the velocity of a particle is also equal to the rate of change of its position w t o r with respect to time. From section 5.3 we therefore conclude that v = % is a tangent vec%or. to the particle path r(t).

The magnitude of the velocity, Iv(t)l, is called the speed of the particle and is often denoted also by v .

The acceleration vector, a( t ) , is defined as the rate of change of velocity with respect to t*~, t . Thus

dv(t) a( t ) = - dt .

Nota t ion

dv(t) d2v(t), . . . may also be denoted by i ( t ) , i ( t ) , . . .. The vectors - - dt ' dt2

Example 3

In each case below, find the velocity and acceleration of the particle whose position vector r ( t ) is given by

108. Khem&tka of r Particle

a r(t)= 2t i + (ta -- 1) j + 3 k b. r f t ) = s i n t i + c o s t j + t 2 k .

Sdution

a. r(t) = 2t i + (t2 - 1) j + 3 k

Differentiating this with respect to t , we have

b. r(t) =s in t i+cos t j + t a k. Then

and

Example 4

If the position vector, r(t), of a particle at time t is given by r(t) = (5 sin 2t - 3) i + 5 cos 2t j show that

a. the .particle moves in a circle in the x y plane

b. the acceleration of the particle is always directed towards the centre of the circle.

Sdution

a. r(t) = (5 sin 2t - 3) i + 5 cos 2t j therefore

x = 5sin2t - 3 , y = 5cos2t, and z = 0.

This is the equauun of a circle in the xy plane with centre (-3,O) and radius 5 units.

b. From r(t) = (5sin2t - 3) i + 5cds2t j

it follows that

dr(t) - 10 cos 2t i - I n sin 2t j v(t) = 7 -

and

= -4(5 sin 2t i + 5 cos 2t j)

Fhm (i) 5 sin 21 i + 5 cos 2t j = r + 3 i. Substituting this into (ii) we obtain

(5 )

(iii) a(t) = -4(r + 3 i).

6.2 Components of Velocity. and Acceleration , . l69). ,

In the diagram below C(-3,O) is the centre of the circular path and P(x, y) is any point on the path.

Hence (iii) becomes a(t) = 4 ~ 2 , that is the acceleration is directed towards C and of magnitude 4 times the radius vector.

Exercises 6.1

In each of problems 1-3 find the velocity and acceleration of the particle whose position vector r ( t ) a t time t is gvien by

1 r( t ) = 3 t a i + 2 t j + t 3 k

2 r(t) = ut cos 6' i + (ut sin 0 - ig t2) j where u, 0, g are constants.

3 r(t) = (1 - e-pt) i + 4 j + (t - 1 + e-pt) k where p is a constant.

4 If the position vector, r(t) , of a particle at time t is given by r( t ) = (2 -sin 3t) i+ (1 + cos 3t) j show that a. the particle moves in a'circle in the xy-plane. . -

b. the acceleration is always directed towards the centre of the circle.

5 The position vector, r ( t ) , of a particle at time t is given by r(t) = sect i + (1 + t an t ) j. a. Show that the particle moves along a hyperbola in the xy-plane. b. Find the velocity of the particle.

6.2 Components of Velocity and Acceleration in some Co-ordinate Systems

The central problem of kinematics is the following: given the initial position and velocity of a body which is subjected to known accelerations, determine the position and velocity of the body a t all subsequent times. In effect therefore, the trajectory of the moving body is to be found for all subsequent times. Some of the trajectories are better described in terms of co-ordinate systems other than the Cartesian system. In this section we derive the components of velocity and acceleration parallel t o the axes of some frequently used co-ordinate systems.

Car tes ian components Since r = z i + y j + x k

170 Kinematics of a Particle

using the properties of differentiation of vectors (see section 5.3) we obtain

and

Thus

(1 ) v = (x,y,i) and a = (x,y,%)

in Cartesian component form.

Radial and transverse components (plane polar co-ordinates) Consider a particle moving in the zy plane. Let its position at time t be P. Suppose its distance from the origin 0 of Cartesian co-ordinate is r, and O P makes an angle t9 with the positive x-axis (see figure 6.2).

Then x = cos B and y = sin 8 so that r* = z* + y* and tan 8 = f . The position of P is also uniquely given by the ordered pair (r,O),r > 0,O 5 O 5 2 ~ , called its plane polar co-ordinates. Let e, be a unit vector in the direction of O> and ee a unit vector perpendicular to e, and in the direction of increasing 8 . Then

0

Therefore de, de, de, dB . -=-s ine i+cosBj=ee and -=- . - - dB dt dB dt -Bee.

/ e

Also

FIG. 6.2

Now r = re, . Therefore dr de, v = - = dt

re, + r- = +e , + r(eee) dt

6.2 Components of Velocity and Acceleration 171

Hence v has radial component i and transverse component re. Also

dv de, . . dee a = - = Fe, + i- + i8ee + roee + r8-

dt dt dt = i: e, + i(8ee) + tiee + reee + re(-ee,)

Hence the acceleration, a, has radial component i: - r g2 and transverse component r e + 2ie. .

R e m a r k 6.1

(i) ee may be obtained by rotating e, anticlockwise through 90°.

(ii) In the fdimensional polar ceordinates, (P , 8, z ) called cylindrical polar co-ordinates, p, 0 are defined as for (r, 8) of plane polar co-ordinates but now with respect to the point Q in the xy plane which is the foot of the perpendicular from the point P(x , y, z ) to that plane (see Figure 6.3).

, ,

X

FIG. 6.3

Tangential and norma l components Suppose a particle is moving along a path C in space. Let the origin of co-ordinates b e . 0 and distance, S, along the path be measured from the point A on C.

FIG. 6.4

Consider two neighbobring points P,Q on C (Figure 6.4) with position vectors r , r + A r respectively and distances s , s + As from A respectively.

A r = and I&(= A s


Hence Ar - ~ s m where is a unit vector parallel to pa. Therefore

ds = - p t

where el is the unit tangent to C a t P. That is

ds Thus the speed v = Ivl of the particle is - so that v = vet.

dt We now restrict ourselves to the case of a particle moving along a path lying in a plane which

we take as the xy plane. At the point P on the path, let the tangent et to C at P make an angle 4 with the positive

z-axis. See Figure 6.5. Let en be the unit vector perpendicular to et and making an angle 5 + 4 with the positive z-axis. en is called the normal to the curve at P. By a derivation similar to that for plane polar co-ordinates e, , ee we have

det - = en den and - d 4

= -et. d 4

FIG. 6.5

By differentiating (4) we have

6.2 Components of Velocity and Acceleration' 173

But

ds where p = - is called the radius of cureraiure of the curve C at P

So d 4

where v = i is the speed of the particle. Thus the acceleration of the particle has tangential component v and normal component d .

9

Example 5

The position vector r(t) of a particle is given by r(t) = 2cost i + 3sintj.

(i) Find when its velocity is perpendicular to r.

(ii) Show that its acceleration is always directed towards the origin.

Solution

r ( t ) = 2 c o s t i + 3 s i n t j

(i) r(t) = -2 sin t i + 3 cost j therefore

r I r ~ r * r = O ~ 5 s i n t c o s t = O e Either sin t = 0 or cos i = 0

n 3n e t = O,n,-, -, . . . + multiples of 2na 2 2

.. d (ii) r = - ( r ) = - 2 c o s t i - 3 s i n t j = - r ,

dl Thus acceleration r is parallel to -r and so is directed towards the origin.

Example 6

A particle moves in' a plane with constant speed show that its velocity is perpendicular to its acceleration

Solution

Let its velocity vector be v(t) at any time t . Since its speed is constant, then Ivl = const. or

v . v = v2 = constant.

Differentiate with respect to time v . v + v . v = o


or

Example 7

The position vector of a particle is given at time t by r ( t ) = 2t i - 3ta j . Calculate v , et and the tangential and normal components of its accleration.

Solution

r ( t ) = 2t i - 3t2 j therefore

(i - 3t j ) Tangential component of a = a . et = 18t

(-6.3 * = d m .

To obtain the normal component of a we first find en = ( a , b) say. Now

en .et = 0 and en .en = I

Thus a - 3 t b = 0 and a 2 + b a = l - - r a = 3 t b and ( 9 t 2 + l ) b = l

so that 1

Since en is obtained by a counterclockwise rotation of et through W (aee the figure above); b > 0 and en = *( 3t i + j ) . Therefore

(3t i + j ) -6 normal component of a = a . en = ( -6 j ) . 4- =

6.3 Motion under Constant Acceleration 175

Exercises 6.2

1 If r(t) = 21 i - 3t2 j, use equation (5) to calculate the tangential and normal components of the acceleration vector. Compare this method with that used in Example 7.

In each of problem 2-4 the position vector r ( t ) of a particle is given. Calculate the speed v , the unit tangent vector et and the tangential and normal components of the acceleration of the particle.

2 r ( t ) = 4 i + t 2 j

3 r ( t ) = ( l - s i n t ) i + ( 2 + c o s t ) j

4 r ( t ) = at i + ( b - d 2 ) j where a , b , c are constants.

5 Starting from x = rcos9, y = rsin9 calculate x and y. By resolving the velocity vector r along e, and ee derive formula (2).

6 A particle moves along a path y = 2cos3x in such a way that t t s component of its velocity in the x-direction is 4 units. Find its acceleration.

6.3 Motion under Constant Acceleration

In this section we consider the motion of a particle under constant acceleration. Let the constant acceleration have magnitude a and direction 8 say. Thus

d v A a ( t ) = a& or - dt

= ae.

Integrating with respect to t we obtain

where c is a constant vector of integration. Suppose at t = 0 that v ( 0 ) = u, called the initial velocity of the particle, then from (i)

u = O % + c * c = u

Therefore

Also dr(t) - a* , + U ~ ( t ) = - -

dt and integrating this with respect to t , yields

1 r ( t ) = -at2 6 + t u + k, k constant.

2

If intially the particle is at the point r ( 0 ) = ro, then in (ii) ro = 0 d + 0 u + k 3 k = ro so that

Now 1

( r - ro) (6 A u) = ( -a t2 8 + t u) . (6 A u) 2

1 - -at2 2 . ( 8 A u ) + t u . ( B A u ) - 2 = o + o = o .


Thus equation (8) represents a plane through ro and with normal e A u. Hence the motion of a particle subject to constant acceleration lies entirely in a plane. The plane is that determined by the acceleration 2, and initial velocity u of the particlesand which passes through the initial position, ro.

We shall take the plane as the xy plane. Take the origin of co-ordinates a t the initid point ro, the y-axis in the direction of 2 and the x-axis perpendicular to 2. Then t = j,u = ul i + or2 j and ro = 0 . Hence equation (8) in component form becomes

giving

x = ult and

For ul # 0 we substitute t = into (9b) to obtain

which is the equation of a parabola with axis parallel to the y-axis, the direction of the acceleration. A motion whose trajectory (or path) is a parabola is called parabolic motion.

For ul = 0, x = 0 for all times t and the motion is along the y-axis only. Such a motion which is along a straight line only is called a rectilinear motion.

We conclude therefore that for a particle moving under constant acceleration, its path is either a parabola or a straight line.

Rectilinear motion In this case the initial velocity and the acceleration are in the same or opposite directions so that UI = 0, u2 = u say in the above. The distance, s, travelled from the initial point and the velocity, v, are given respectively by

( l la . )

( l lb .)

The formulas,

1 2 s = ut + -at 2

from (8)

v = u + a t from (7)

are easily derived from (11) for:

1 1 1 s = -t(2u + at) = - t (u + u + at) = + v ) 2 2

and

which simplifies to v2 = u2 + 2as.

6.3 MotionunderConstantAcceleration 177

Parabolic motion In this case x = ul t and y = $at2 + u2t. l+om (7)

v( t ) = at j + ul i + u2 j * V I = UI (constant) and v2 = u2 + at .

Thus there is constant velocity along the x-axis, that is perpendicular to the constant acceleration.

Remark 6.1

For rectilinear motion, since only two integrations are performed only two of the four equations (11) and (12) are different: the others are just ways of rewriting the two. Hence in any problem only a t most two of these equations need be used. The idea is t o determine which would yield the solution easiest.

Example 8

A small stone which is thrown vertically upwards with an initial speed of 80 m/sec is acted upon by an acceleration (called gravity) of 10 m/sec2 in the downward direction. Find

(i) the maximum height above the point of projection which is attained by the stone;

(ii) when the particle is a t a height of 140 m above its point of projection.

Solution

Here the initial velocity and acceleration are in opposite directions. Hence the motion is in a straight line. Take the origin as the point of projection and positive x-axis vertically upwards. Then

x = -10 Integrate with respect to t

(*I x = -lOt + c

At t = 0, x = 80, therefore 80 = -10(0) + c * c = 80

i.e. x = 80 - lot Integrating again,

(**I x = 80t - 5t2 + k

B u t a t t = 0 , x = O ~ O = O + O + k ~ k = O P- 0 1

Therefore x = 80t - 5t2 (***I (i) At maximum height x = 0 * 80 - lot = 0 * t = 8 sec. Therefore maximum height = 80(8) - 5(8)2 = 320 m. (ii) The particle is a t a height of 140 m above 0 if x = 140. Tha t is

Therefore t = 2 Qr 14 Note that when

is moving down (x = acceleration equation

sec. t = 2, the particle is moving up (x = 80 - 20 = 60) and a t t = 14 the particle 80 - 140 = -60). We have used the alternative method of writing down the and integrating step by step instead of just quoting the formulae ( l l ) , (12).

Example 9

A particle which is projected a t an angle of 60° to the horizontal with speed 100 m/sec moves freely under gravity. Determine (i) its maximum height (ii) its range on a horizontal plane through


the point of projection and (iii) when its velocity is inclined at an angle of 45O to the downward vertical. [Take acceleration due to gravity, g, as 10 m/sec2].

Solution

The acceleration is comtant vertically downwards of magnitude 10 m/sec2. The initial velocity however has both vertical and horizontal components. Therefore we have parabolic motion. Take the y-axis vertical with positive values upwards, the x-axis horizontal and the origin a t the point of projection.

Vertical and horizontal component equations of acceleration are

ji=-10 and x = O

respectively. Integrating these with respect to t gives

3i=-10t+cl and z = c 2 .

Initially a t t = 0,$ = 100sin 60 = 100 . 9 = 50& and i = 100 cos 60 = 100 . $ = 50. Hence substituting we obtain

= 5 0 6 - 102 and x = 50.

Integrating again y = 5 0 h t - 5t2 + dl and x = 50t + d2

Since origin is at point of projection dl = 0, d2 = 0 giving

y = 5 0 h t - 5t2 and z = 50t.

(i) At maximum height vertical component of velocity is zero.

Therefore Maximum height = 5 0 4 5 4 ) - 5 ( 5 6 ) ' = 375m.

(ii) The particle is on the level of projection again when y = 0

6.4 Motion under Variable Acceleration 179

Therefore t = 0 (at origin) or 1 0 h sec.

Therefore Range R, which is the distance travelled on the ground

(iii) At Q the velocity is inclined at an angle 4 to the downward vertical when y < 0, x > 0 and x x

tan 4 = - (see figure) = - I Y I (-Y). For 4 = 45" we have

Exercises 6.3

1 A particle is projected vertically upwards from the edge of a 30 metre cliff with an initial speed of 120 m/sec. Find (i) the maximum height above the cliff attained by the particle. (ii) the time taken by the particle to hit the bottom of the cliff. (iii) the speed with which the particle hits the bottom of the cliff. Take the acceleration due to gravity as 10 m/sec2 in the downward direction.

2 A particle is projected at an angle of 45" to the horizontal with speed 150 m/sec and moves freely under gravity. Determine its maximum height and its range on a horizontal plane through the point of projection. Take acceleration due to gravity as 10 m/sec2.

6.4 Motion under Variable Acceleration

Recall the central problem of kinematics of a particle, namely, given the acceleration of a particle and its initial position and velocity, to find its position at all subsequent times. In general this acceleration, a, may depend on t , r and r so that a = f (t, r, r). The integration of this equation to obtain r(t) is not possible for arbitrary f . In the following subsections we consider cases for which the integration is within the scope of a first year university calculus course.

6.4.1 Rect i l inear mot ion

Let the line of motion of the particle be the x-axis. The acceleration x may be given as a function of time t , displacement, x, or speed v .

Case 1: a = f ( t ) 1.e.

dv - = f (t) ~ ( r ) = f (A) dA + constant. dt

180 ' Kinematics of a Particle

Setting t = 0 yields

Thus

= g ( t ) + v ( 0 ) say.

Integrating again,

Case 11: a = f ( x ) - , d; dv dx dv

Note that a = - = -. - = v- so tha t dt d x dt d x

Integrating with respect to x, we have

F'rom the above we can get v as a function of x, h ( x ) say. Thus

Integrating again with respect to x yields

which gives x as an implicit function of t .

Case 111: a. = f ( v ) . dv

Here as in case I1 we set a = v- so that d x

which may be integrated with respect to v to obtain


We need to solve for v explicitly as a function of x and thereafter proceed as in the second integration in Case I1 to obtain x as a function o f t .

In the alternative we use

dt 1 which implies - = - . Thus integrating with respect to v, we obtain

dv f(v)

Once again we need to solve for v explicitly as a function o f t and thereafter proceed as in Case I.

Example 10

The acceleration of a particle moving in a straight line is given at time t by a(t) = 2t - 3t2. At time zero the particle is at the origin and has zero velocity, Find

(i) the farthest distance the particle is from its starting point before returning there.

(ii) when it returns to its starting point.

Solution

Let the line along which the particle moves be the x-axis with origin at starting point.

Integrating with respect to t , k t 2 - t 3 + C 1 ,

Now x(0) = 0 cl = 0, therefore i = t 2 - t3.

Integrating again, we have 1 1

x = -t3 - -t4 + C2. 3 4

Using x(0) = 0, we obtain c2 = 0, so that

(i) At farthest point from starting point x = 0 i.e.

t = 0 gives the initial time. Hence

(ii) The particle returns to its starting point when x = 0 i.e.


Hence particle returns after time

Example 11

In one of the ran the first acceleration,

4 units.

U

heats for the 1500 m race during the centennial Olympiad (Atlanta '96) a competitor 100 m at constant acceleration, the last 200 m in 23 sec a t a (different) constant

and finished the race at a speed of 10.6 m/sec. If he covered the remaining distance at a constant speed, find

a. his initial constant acceleration,

b. the final constant acceleration,

c. his uniform speed, and

d. his total time for the race.

Solution

FIG. 6.6

Figure 6.6 is a velocity-time graph for the motion of the competitor. Since v = u + at for constant acceleration, segments of the race run a t constant acceleration are represented by straight lines on the graph. O A represents the initial 100 metres covered at constant acceleration, say, a l (# 0), AB the segment covered a t uniform speed say w , and hence zero acceleration, and BC the last 200 m covered at uniform acceleration, say, as(# 0). Let t l sec, t 2 sec be the times for running the first and second segments respectively.

Recall the equations governing motion under constant acceleration: v = u + at , 6 = ut + i a t 2 , v 2 = u2 + 2as and s = $(u + v)t in the usual notation.

le t segment Initial velocity = 0, final velocity = w, distance travelled = 100 m, acceleration a l , time t l .

w = O+al t l

w2 = 0 + 2a1(100)

2nd segment No acceleration, speed constant at w m/sec. distance travelled = 1500 - 100 - 200 = 1200 m. Therefore

1200 = wt2

3rd segment

(i) (ii)

(iii)


Acceleration as, initial velocity w, final velocity = 10.6, distance covered = 200 m, time taken = 23 Bec.

These are 5 equations for the five unknowns a l l a3,tl,t2 and w. Solving [for example w from (v), then a3 from (iv), t2 from (iii), a1 from (ii) and t l from (i)] we obtain a. a, = 0.23m/sec2 b. as=O.l7m/sec 2

c. w = 6.79m/sec d. total time = t l + tz + t3 = 29.53 + 17.73 + 23 = 229.3 sec or 3 min + 49.3 sec.

Example 12

Derive the formula v2 = u2 +2as for linear motion with constant acceleration a by direct integration.

Solution

To introduce distance x we use the device in cases I1 and I11 of writing

when acceleration is f (x) or f (v). Thus dv

v- = a dx

Integrate with respect to x, 1 -v2 = ax + c 2

Initially x = 0 when v = u which implies

Therefore 1 1 -v2 = ax + -u2 or v2 = u2 + 205, 2 2

Example 13

A particle moves in a straight line in such a way that x = -w2x where x is its displacement from a fixed point 0 on the line.

a. Find an expression for z(t)

b. If w = 3 and the particle is initially at reet 5 unite from 0, find


(i) its speed when it is 3 units from 0; and

(ii) how long it takes to reach 0 for the first time.

Solution

a. 2 x=-W x

dv dv dx dv ~ ~ t $ = - = - - - - v-. Thus dt dx dt dx

Integrating with respect to x, we have

Since v2 2 0, then -;w2x2 + c 2 0 also, so that c 2 ;w2x2 >_ 0, Thus c is a nonnegative number (it must actually be positive unless x = 0 for all time, which is a trivial case). Taking c = io2a2 we have

Therefore v = fwd-. We now arbitrarily select the positive sign, so that

Integrating with respect to x gives 1 f dx

To evaluate the integral on the right, put x = a sin 6 3 dx = a cos6 d6 and

Thus 1

t = -8 + k, k a constant or 6 = wt - wk = wt + a say, W

Now 2

sin 6 = - = sin(wt + a ) or x(t) = asin(wt + a) . a ( W

The right hand side of (IV) may be expanded to obtain

Equations (IV), (V) are alternative forms of the solution of equation (I). In each case there are two constants of integration, a , a and A, B respectively, to be determined using the initial conditions for the particle in any specified problem. These equations are called general solutions of ( I ) .


dx Show that using v = - = -w Js also leads to (V) so that the solution of (I) is

dt independent of the choice of positive or negative sign. b. Here x = a sin (3t + a ) since w = 3 . When t = 0 , x = 5 and x = 0 . Therefore

x(0 ) = 5 = a sin a and x(0 ) = 0 = 3a cos a

which imply that n lr

cosa = 0 or a = - and 5 = asin- = a. 2 2

Therefore l r x ( t ) = 5 sin (3t + -) = 5 cos 32.

2

(i) Speed v at any time is given by

Therefore speed v when x = 3 satisfies v2 = 9(52 - 3 2 ) = 9(16) and hence v = 12. (ii) Particle reaches 0 when z = 0

eo that n nlr t = - + -

6 3 n

and the smallest value o f t is - 6 '

Example 14

The acceleration of a particle moving in a straight line is given by -v2 where v is the speed of the particle. If initially the particle is at the origin and has speed v = 10, determine the position of the particle for all time.

Solution

Integrating with respect to v , we have

1 t = - + constant

v

1 1 When t = 0 , v = 10 * 0 = - + constant constant = -- Thus

10 10'

Hence 10

v = - d z 10

1 + lo t which implies - = -

cEt 1 + 1 o t '

186 Kinerna?&:s of a 1 a -tide

Integrating with respect to 1, we have

Alternatively dv 2 dv

Use v- = -v or - = -v so that dx dx

ds 1

Integrating with respect to v , yields

where A is a constant (= e t c ) . When x = O , v = 10- 1 0 = A . 1 . Hence

Integrating with respect to x yields 1

t = -ez + B. 10

1 1 1 t = O 1 ~ = O ~ O = - . 1 + B ~ B = - -

10 10' Hence t = -(ez - 1). Thus

10

6.4.2 Motion in a plane

For the case of central orbits, the acceleration of the particle in space is always directed along the line of the radius vector from the origin to the particle. That is,

where f (r) is some function of r = Jr(. Taking the cross product of r with (i) gives

Note that

(ii)

d 'Therefore (ii) implies -(r A i-) = 0. Integrating with repect to t we obtain

dt

(iii)


where h is a constant of integration. Taking the scalar product of (iii) with r yields

which is the equation of a plane containing the origin and with normal h. F'rom (iii) h is perpendicular to both r(t), i ( t ) for all times and in particular fort = 0. Hence a particle subjected to an acceleration always along its radial vector moves in a plane defined by its initial position vector, ro, and velocity u.

Example 15

A particle mova in such a way that its acceleration is given by r = -w2r . If initially the particle is at a point A distant a from the origin of co-ordinates 0 and has speed u perpendicular to 02, find the path of the particle.

Solution

The acceleration is directed towards the origin. Hence the motion of the particle lies in a plane containing the origin. Take this plane to be the xy plane. Take OA as the positive x-axis and the positive y-axis in the direction of the initial velocity. Then

2 x = -w2x and y = -w y

By Example 13, these equations have general solutions

x(t) = A sin wt + B cos wt and y(t) = Gsin wt + H cos wt

respectively. But at t = 0, x = a , y = 0 and x = 0, y = u,

Thus x(O)=a=AsinO+BcosO=+ B = a

x=O=AwcosO-Bsinw*A=O

Also y(0) = 0 = GsinO+ HcosO =+ H = 0 u

~ ( 0 ) = u = G w c o s O - H w s i n O a G = - W

u Therefore x(t) = a cos wt and y(t) = -sin wt

W

which is the equation of an ellipse or circle if u = aw.

Example 16

The radial and transverse components of the velohity of a particle moving in a plane are given by krn, p / r2 respectively. Find

(i) the equation of the path of the particle if it passes through the point (r, 0) = (1, O),

(ii) the radial and transverse components of the particle acceleration.


Solution

By section 6.2, the velocity v of the particle is v = i e , + roee in polar co-ordinates where g , e e are unit vectors in the radial and transverse directions respectively. Therefore

' P i = krn and T O = .

dr (i) For the equation of the path we need to obtain - and solve for r (0) .

do

Therefore do p 1 -- - -- do or k- = p-("+'). dr k rn+3 dr

Integrating with respect to r gives

= p T - ( ~ " ) + constant, n + 2 0 n + 2

But when r = l , d = 0 therefore

I' O = ,u (2) + constant a constant = - n + 2 n f 2 '

Thus

(ii) a = (i; - r e2 ) e, + ( r e + 2i9) ee. Now

Therefore radial component of acceleration, a,, is

Now

Therefore transverse component of acceleration, ae, is


Example 17

A particle moves in a plane with constant sped v . If the magnitude of its acceleration is v a / ( s a + 4) where s is its distance along its path of motion, obtain an expression for s.

Solution

For a particle moving in a plane its velocity v and acceleration a are

respectively where e t , en are unit vectors in the tangential and normal directions respectively (see eection 6.2).

d' Since v = constant, d = 0 and a reduces to a = va- en. Therefore ds

Integrating with respect to s, we have

where cr is a constant of integration. Therefore

Exercises 6.4

1 The acceleration of a particle moving in a straight line at time t is given by a(t) = 10 - 2t. At time zero the particle is at the origin and has zero velocity. (i) Find the farthest distance the particle is from its starting point before returning there. (ii) When does it return to its starting point? (iii) What is its speed when it returns to its starting point?

2 The acceleration of a particle moving in a straight line is the negative of its velocity. If initially the particle is at the origin and has speed v = 25, determine the subsequent position of the particle.

Kinematics of a Particle

1 The acceleration of a particle moving in a straight line is -2 where x is its distance from a - fixed point. If initially the particle is at x = 1 and has epeed fi, find its positlon as a fumtion of time, t. The radial and transverse components of the velocity of a particle moving in a plane are given respectively by kr2, pr-' . Find (i) the equation of the path of the particle if it passes through the point (r, 8) = (1, O), (ii) the radial and transverse components of the particle acceleration.

In a 100-metre race on a straight track the acceleration a(t) in m/sec2 of an athlete is given by

4 O S t 5 3 0 3 < t 5 1 0

-12 10 < t 5 11 Find (i) the maximum speed of the athlete, (ii) how long the athlete takes to complete the race, (iii) the speed of the athlete as he breasts the tape. (iv) how much further the athlete ran before stopping.

The acceleration of a particle is -a2z i - 4 ~ r ~ ~ j . Its initial position and velocity are r = (0, b) and r = (u, 0) respectively. (i) Find the position of the particle as a function of time t .

2a2 b (ii) Show that the path of the particle is the parabola y = b - - x2.

u2 The relativistic equation of acceleration of a particle moving in a straight line is given by

where m, c are constants. Find the distance travelled from rest in time t for F constant.


If the position vector, r(t), of a particle at time t is given by r(t) = (1 - 2sint) i + (2 + cost) j a. show that the path of the particle is elliptic and b. find the velocity and acceleration of the particle. If the position vector, r(t), of a particle at time t is given by r(t) = 3sint i + 3 cost j + 4k show that a. the path of the particle is a circle; b. the particle travels at constant speed; and c. the acceleration of the particle is always directed towards the centre of the circle. The position vector of a particle at time t is given by r(t) = sint i + 2costj. Calculate the speed v, the unit tangent vector to the path of the particle ei, the unit normal vector en and the tangential and normal components of its acceleration. Starting from x = r cos 8 , y = r sin 8, calculate x and y. By resolving the acceleration vector along e, and ee obtain formula (3). A particle moves along a path y = Asin wx in such a way that the component of ite velocity in the x-direction is u where A, w and u are constants. Find its acceleration.


6 In an 800-metre race on a straight track the acceleration a(t) in m/seca of an athlete is given by

O < t < 3 0 0 3 0 < t 5 9 0 -+ 90 < i _< 110

Find (i) the maximum speid of the athlete, (ii) how long the athlete takes to complete the race (800 metres), (iii) the speed of the athlete as he breasts the tape.

7 The acceleration a(t) of a particle is -4a2xi - aZyj. The particle is initally at rest at the point (a, b). (i) Show that the particle oscillates along an arc of the parabola 2ay2 = b2(x + a). (ii) Find the velocity of the particle whenever it is at the points (a, -b) and (-a, 0).

8 The radial and transverse components of the velocity of a particle moving in a plane are given respectively by krn , prm , rn # n. Find (i) the equation of the path of the particle if it through the point (r, 0) = (1, O), (ii) the radial and transverse components of the particle acceleration.

9 A particle is projected with speed u at an angle 0 above the horizon. For each of the cases below, determine the time interval during which the particle is at least at a height h above its point of projection. a. u = 80 m/sec, 8 = 60°, h = 60fi rn. Take g = 10 m/sec2) a. u = 50 m/sec, 8 = 90°, h = 10 m. Take g = 9.8 m/seca)

10 A particle is projected with speed u at an angle of elevation of 60°. Determine in terms of u and g when and where its velocity is a. inclined to the upwards vertical at an angle of 30° b. perpendicular to its initial velocity.

11 By using vector methods, or otherwise, show that the time elapsed before a projectile is moving perpendicular to its initial direction of projection (as in problem 1Ob above) is the same for all. angles of projection.

12 A bullet is projected from a point A with speed 40 mJsec at an angle of 60° towards another point B on the same horizontal level. Simultaneously, a second bullet ia also projected from B towards A with speed 80 m/sec at an angle of 45O. If the distance A B '= 100 m and the two bullets lie in the vertical plane containing AB, a. find the height of the first bullet when it is vertically above B; b. a t what instant are the two bullets in the same vertical lane; and c: what is then the distance between them? (Take g = 10 m/sec2).

Dynamics of a Particle In the chapter on the kinematics of a particle we discussed the motion of a particle subjected to specified accelerations. In this chapter we try to determine the sources (called forces) of these accelerations. The mathematical model of the motion of a particle is based on Newton's three laws of motion one of which relates accelerations to the forces which produce them. The model is called Newtonian (model of) dynamics.

7.1 Newton's Laws of Motion

1s t Law Every object continues in its position of rest or state of uniform motion in a straight line unless acted on by an external impressed force.

2nd Law The rate of change of momentum of an object is proportional to the external impressed force, F.

DEFINITION. The momentum, p , of a particle is the product of its mass m and velocity, v ; that is, p = mv.

3rd Law If one particle exerts a force on another particle, then the second particle also exerts a force on the first; these forces are equal in magnitude, opposite in direction and act along the line joining the two particles.

This third law is often stated as "action and reaction are equal and opposite". Newton's first law prescribes what the normallstandard state of motion of an object is, namely,

the state of rest or of unzform motion in a straight line. This state is also called eqlrilibrium etate. Any deviation from this implies the existence of a force acting on the object. Note that the condition of rest is stated explicitly merely as emphasis since it may be taken to represent uniform straight line motion with speed of zero!

From the second law we have

d - (mv) oc F dt

where k is a constant. Usually by choice of units of force F, k is taken to be unity. Thus

7.1 Newton'sLawsofMotion 193

dm If the particle has constant mass then - = 0 and (2) reduces to

dl

Example 1

A mass of 6 kg is placed on a horizontal shelf which ascends with

a. an acceleration of 5 m/sec2

b. uniform speed of 3 m/sec.

Find the force on the shelf due to the mass in each case.

Solution

a. By Newton's 3rd law, the force on the shelf = the force on the mass.

The force on the mass = mass x acceleration

= 6 x 5kg.m.s-2

= 30N

b. . Since the speed is uniform, the acceleration is zero. Therefore

The force on the mass = mass x acceleration

= 6 x 0 = 0.

Applications Applications of Newton's 3rd law are many. We mention briefly here two broad areas of application. 1. Objects (e.g. canoes, ships, submarines) which move on or in water do so by having the water pushed backwards (i.e. action). The reaction of the water on these objects move them forward. 2. Jet propulsion in the air is achieved by pushing out a stream of hot gases from the jet engines. The escaping gases produce a reaction on the aircraft which pushes it forward. A simple demon- stration of this principle is to inflate a toy balloon, and then release it with its mouth open. The escaping air causes the balloon to zigzag through the air!

The rest of the chapter will be devoted to illustrative examples of the use of these laws to solve specific problems.

Notation

In diagrams we shall indicate forces by a single arrow +, and acceleration by two arrows -+.

Exercises 7.1

1 A 1000 kg car moving at a speed of 80 km/hr is brought to rest in 200 m by a constant braking force. Determine (i) the acceleration of the car (ii) the constant braking force.

2 Suppose a body of mass 15 g which is initally a t rest is subjected t o a constant force. If i t attains a velocity of 40 m/sec within a distance of 60 m find the force acting on it.

7.2 Pulleys; Motion on Smooth Surfaces

Example 2

A particle of mass rn moves from rest on a smooth plane inclined to the horizontal at an angle a. Determine the force exerted on the particle by the plane and the motion of the particle on the plane.

Solution , ,

FIG. 7.1

When the particle is a t any point P on the plane there are only two forces acting on it. These are the force of gravity mg vertically downwards and the force exerted by the plane (called reaction of the plane, R, on the particle and this acts along the perpendicular to the plane as it prevents the movement of the particle in that direction (see Figure 7.1).

If we take the positive z-axis'down the plane and the positive y-axis perpendicular to the plane and directed upwards then the components of the equatioq of motion in these directions are respectively

mx = mg sin a

my = R - mg cos a. (i)

(ii)

However the particle remains on the plane throughout the motion so that y ( t ) = 0 for all 2 . Thus y(t) and y(t) are both zero. Hence equation (ii) becomes

Equation (i) is equivalent to x = g sin a

which implies that the particle slides down the plane at constant acceleration a = g s ina .

Remark 7.1

The word 'smooth' means that the reaction force between the particle and plane has no component (called fictional force) along the plane. F'riction normally opposes ttctual or potential motion.

7.2 Pulleys; Motion on Smooth Surfaces 196

Pulley Sys tems Pulleys are useful devices which enable us lift/lower heavy loads by using numerically smalles- fvrces t h a the weight of the load. The basic component of a pulley system consists of a grooved wheel over which a rope in tension passes. (In Figure 7.2 this tension is produced by weights attached to the ends of the string). A pulley system can consist of many such components.

In real life the string moves without sliding over the wheel and thus causes the pulley to rotate about its axis, 0. The tensions in the two halves of the string must be

B -:: of unequal the pulley since about they provide its axis. the This motive is no force longer for the motion rotation of a

particle and we must wait till we learn the theory of motion A of large bodies before tackling such problems. However,

the assumption of a frictionless pulley greatly simplifies the problem and enables us get (approximatel results about

FIG. 7.2 the working of such pulley systems. If the pulley is smooth then the string slides over the pulley without turning it and so the wheel

need not rotate a t all. Suppose in Figure 7.2 that the weight A, B are distances x, y respectively below the origin 0.

Since the string is inextensible, its length, 1, is constant and so

x + y + .Ira = I, constant.

Differentiating successively with respect to t , we get

and

Thus the velocities and accelerations of A and B are respectively equal in magnitude but opposite in direction. .

Example 3

Two particles of mass ml and mz respectively are connected by a light inextensible string which passes over a fixed smooth pulley. Obtain the acceleration of the particles and the tension in the string.

Solution

Since the pulley ie smooth, the tendons in both parts of the string are equal. Suppose the ml mass moves downwards with acceleration a, then the m2 mass movee upwards with the same acceleration.

196 Dynamics of a Particle

Forces acting on the particle A of mass ml are T, tension in the string and gravity mlg (see Fig. 7.3). Therefore equation of motion of A is

mla = mlg - T. 6 )

Forces acting on the m2 mass B are T, tension in the string and gravity m2g. Therefore equation of motion of B is

m2a = T - m2g. (ii)

Adding (i) and (ii):

(ml + m2)a = mlg - m2g a = d m 1 - m2) , (iii) fw? m l + m2

FIG. 7.3 Eliminating a from (i) and (ii) or using (iii) in (i) say, gives

Example 4

A particle of mass m is at rest on the top of a fixed smooth cylinder of radius a. It is given a small velocity u in the plane perpendicular to the axis of the cylinder so as to slide down the side of the cylinder. Show that the particle will leave the surface of the cylinder. Find where it leaves the cylinder and its velocity at that point.

4

Solution

At time t let the particle be at the point P on the cylinder. Take the cross-section of the cylinder perpendicular to its axis and containing the point P: this is a circle. Let 0 be the centre of this circle. The force8 acting on the particle are (i) its weight mg vertically downwards and (ii) the reaction, R, of the cylinder on the particle in the normal direction, i.e. along the radial vector 03. The equation of motion of the particle is then

7.2 Pulleys; Motion on Smooth Surfaces 197

where e is a unit normal in the vertical direction.

FIG. 7.4

Take the vertical OA through 0 as positive x-axis, the horizontal through 0 in the direction of increasing 6 as the positive y-axis where 6 is the angle that O P makes with the positive z-axis (Figure 7.4). Take the z-axis to complete a right-handed ceordinate system. Thus

R = R(cos 6i + sin Oj), e = -i, r(0) = u j and r(0) = a i.

The third component of equation (i) is q t ) = 0.

Integrating with respect to t , this yields f ( t ) = cl. Since f(0) = 0, then 0 = cl and E(t) = 0. Integrating again we have

%(t) = c2.

But z(t) = 0 =j cz = 0 so that z(t) = 0 for all t . Thus the motion lies entirly in the xy plane, the vertical plane containing the initial displacement and velocity. Now use polar ceordinates. Since r = Ir\.= constant, i = 0 and the radial and transverse components of acceleration reduce to -ae! and a0 respectively. Also the components of the velocity in these directions reduce to 0 and a6 reepectively confirming that all motion is tangential to the radius vector so that v = a0.

Resolve the equation of motion (i) along the radical and transverse directions we obtain

1 -ae2 = -R - g cos6

m (ii)

a6 = g sin 6 (iii)

.. de de d6 .dB ... Y Using 6 = - = - - = 6-, (111) becomes

dt do dt do

. dB a6- = g sin 8.

do

Integrate with respect to 0 to obtain

1 . -ae2 = -g cos 6 + constant 2

At t = 0, a6 = u, 6 = 0, therefore

1 2 1 ua -p (:) = -1 1 + constant =+ constant = -- + g. 2 a


1 . 1 u 2 Hence -a02 = -- + g - g cos 0 or

2 2 a

a2@ = u2 + 2ag(1- cos 8)

Substituting (iv) into (ii) yields

1 = g cos 8 - - [u2 + 2ag(l- cos O)]

a

The particle remains on cylinder as long as R > 0, it leaves it if R = 0. Thus

This is an acute angle since cosO > 0. So particle leaves cylinder at a = cos-' ( u 2 iiag) where

0 5 a < $. However, since I cos Ol 5 1, this is a real angle if

If ua 2 ag, particle leaves cylinder immediately on projection. To find velocity when particle leaves cylinder, we use (iv)

Exercises 7.2

1 Two masses 120 kg and 90 kg are connected by a light inextensible string passing over a smooth fixed peg. .Find the distance, from rest, that they will describe in 4 sec and the speed acquired by either mass in that time. Take the acceleration due to gravity as 10 m/sec2.

2 Two masses ml , mz are connected by a light inextensible string passing over a light smooth pulley. If the pulley moves with an upward acceleration equal to that of gravity, g, show that

the tension of the string is 4mimzg m1+ m2

3 An object of mass 5 kg slides from rest down a 60°-incline of length 60 m from the top. Neglecting friction and taking the acceleration due to gravity as 10 m/sec2 a. how long will it take to reach the bottom of the incline? b. what is the speed with which it reaches the bottom?

7.3 Projectile Motion 199

4 A particle of mass rn slides without rolling down a frictionless inclined plane AB of angle a and length I . If it starts from rest at the top A of the incline, find (i) the ,acceleration (ii) the velocity, and (iii) the distance travelled after time t . (iv) the time taken by the particle to reach the bottom B and its speed at B.

7.3 Projectile Motion

In this section we study the motion of particles projected into space and moving under gravity. Consider a particle of mass m projected with velocity u which makes an angle 0 with the

horizontal plane. If we neglect the effect of air resistance then the only force acting on the particle is that due to gravity. Hence its equation of motion is

where 6 is a unit vector in the upward vertical direction. Since g is constant, we have the case of motion under constant acceleration and by section 6.3 this motion lies in a plane which we take as t,he xy plane.

Take origin 0 at the point of projection, the positive y-axis as vertically upwards and the positive x-axis horizontal so that 0 5 90' (see Figure 7.5). q

FIG. 7.5

The horizontal and vertical components of equation (i) are

x = O and y = - g .

Integrating with respect to t , we have

x = cl and ~ = -gt + c2 But at t = 0, x = u cos 0, y = usin 0 therefore

c l = u c o s e and u s i n e = c 2

(ii)


Hence x=ucos f l ' a n d y=us in f l -g t .

Integrating once more yields

1 x = u t c o s f l + d l and y = ~ t s i n f l - - ~ t ~ + d ~ .

2

Since x(0) = y(0) = 0, dl = 0 = d2 so that

1 x = ut cos 19 and y = ut sin 8 - -gt2

2

The path of the particle is obtained by eliminating t between x and y, i.e.

2 1 2 2 y = usin0 (-) ucose - (-) u cos e

or 9 y = x tan 8 - -(I+ tan2 fl)x2.

2u2

(iii)

This is a parabola which passes through the origin.

Maximum height The particle attains its maximum height H when its vertical component of velocity 2j = 0. That is

u sin fl usinfl - gt = 0 * t* = -

9

where t* is time to maximum height. Therefore

From the above expression we see that the overall maximum H,,, for all 8 is obtained when sin 8 = 1, that ia 19 = f. Thus H,, = $ when the particle is projected vertically upwards.

Range on horizontal plane of projection The particle is on the horizontal plane through the point of projection when its vertical height y is zero. That is

1 y = ut sin fl - -gt2 = 0

2

1 2u sin fl *t(usinfl- -gt)=O-t = O or -.

2 9

Therefore 2u sin fl 2u sin fl

Ftange R = (T) = ucosfl . - 9

2u2 u2 = - sin 19 cos 19 = - sin 28.

9 9

Clearly the angle of projection 8 to obtain ,maximum possible range R satisfies sin 28 = 1 =$ 20 = % o r e = $ .

Therefore

Note that since time to return to level of projection is - which equals twice the time to maximum height, then the time for upward motion equals the time for downward motion.

Hitting a target We wish to obtain conditions on u, 8 for a projectile to hit a target with co-ordinates (a , b) say. The point (a, b) should lie on the trajectory (v) above, that is

9 b = a tan 8 - - ( I+ tan2 8)a2. 2u2 (vi)

a. If 8 is specified, what will be the value of u for hitting target? From (vi).

u = a J 2(a g ( l + t a n 2 e ) tan 8 - b) or asece JT 2(a tan 8 - (vii)

g( l + tan2 @)a2 Since u2 2 0, then

2(a tan 8 - b) 2 0

Thus if 8 is given, subject to 0 > tan-' !, the required speed of projection is given by (vii). b. If u is specified, what will be the value of 8 for hitting target? If a # 0, equation (vi) is a quadratic in tan 8 giving the two possible angles of projection to hit the target. It may be rewritten

(viii)

For real valuea of tan 8, the discriminant must be non-negative;

Completing squares in u2 we have

Taking square roots we have


However

Therefore u2 - bg > 0 so that lu2 - bgl = u2 - bgr giving u2 - bg 2 g m or

This is the minimum initial speed for hitting the target and tancu for the angle a of projection is then given by the solution of (viii).

Example 5

A particle is projected with velocity 30 m/sec in a direction making an angle 30° with the horizontal. Find its position and speed after 2 seconds. Take g = 9.8m/sec2.

Solution

FIG. 7.6

Take x-axis horizontal and y-axis vertical (see Fig. 7.6). Let the particle be at A after 2 sec. Vertical motion:

i ( t ) = u sin a - gt = 30 sin 30' - 9.81

= 15 - 9.8t

Therefore at t = 2, y(2) = 15 - 9.8 x 2 = -4.6 msec, y(2) = 15(2) - 4.9(2)2 = 10.4 m. Horizontal motion:

4 x = ucoscr = 30cos60~ = 30. - = 1 5 6 2

x(t) = 1 5 6 t

Therefore 4 2 ) = 15&(2) = 3 0 a . Hence co-ordinates of A are (x(2), y(2)) = ( 3 0 6 , 10.4). Also

Therefore v(2) = 26.4 m/sec.

Example 6 I . .'.

A particle is projected with speed 2 m m/sec directly towards a wall 20 m away. If it just clears the top of the wall which is 5 m above the horizontal plane through the point of projection, find two possible angles of projection. Take g = 9 . 8 m ~ - ~ .

Solution

FIG. 7.7

Fig. 7.7 illustrates the path of the pa.rticle. Since the particle just clears the top of the using

wall, then it passes through the point (20,5). Hence

we have

= 20 tan a - 7(1+ tan2 a )

7 t a n 2 a - 2 0 t a n a + 1 2 = 0

( 7 t a n a - 6)( tana - 2) = 0

Therefore 6 6

t a n a = - or 2 a = tan-' - or tan-' 2. 7 7

Example 7

A bird is flying with velocity of 14 m.sec-l at a constant angle of elevation of 60°. At the instant' when it is 10 m vertically above a boy, the latter throws a stone a t it a t an initial angle of a to'the horizontal. Show that for the stone t o hit the bird t a n a 2 2 + 4. (Take g = 9.8 m . ~ e c - ~ )


Solution

FIG. 7.8

Take the origin as the point off projection, the positive y-axis vertically upwards and the positive z-axis horizontal so that Oxy is in the plane of projection of the stone. See Figure 7.8. Let stone be projected with speed u a t an angle a to the positive x-axis. For motion of bird:

xb(t) = 14 cos 60 t = 7t

ya(t) = 10 + 14 sin 60t = 10 + 7 6 1

For motion of stone:

x,(t) = ucosa t

1 2 y, (t) = u sin a t - Igt

Stone hits bird e rb(t*) = r,(t*) for some t* > 0.

Thus 1

7t* = u cos a t* and 10 + 7 6 t ' = usin a t * - -gt*2 2

which implies .

7 = ucosa (i) 1

10 + 7 4 t * = usin a t * - -gt*2 2

( ii)

These are 2 equations in the 3 unknowns t* ,u , and a. The best we can achieve is to eleirninate one of the variables between the 2 equations.. The answer requires a condition on t a n a so that a should be retained. Also t* appears only in one equation and thus cannot be eliminated. Hence we eliminate u.


Substitute for u from (i) into (ii)

1 -gt*2 + ( 7 4 - 7 tan a) t* + 10 = 0 2 4 W 2 + 7 ( 4 - tan a) t* + 10 = 0

49P2 + 7 0 ( d - tan a ) t * + 100 = 0

Since product of roots = = '00 > 0, the roots are either both positive or both negative. We 49.

therefore need, two positive roots since t* > 0. But

-TO(& - tan a ) f J [ 7 0 ( 8 - tan all2 - 4(49)100 t* =

2.49

This is of form t* = X f dm. TO have any positive root at all, X must be positive

For real roots X 2 - p2 2 0

That is t a n a - 4 2 2 or t a n a 2 2 + f i .

Range on an inclined plane Suppose now that the ground is not horizontal but is inclined at an angle a to the horizontal. Let a particle be projected up the plane with speed u from a point 0 on the plane in a direction making an angle 6 with the horizontal so as to move in the vertical plane containing the line of greatest slope of the inclined plane (see Figure 7.9). Using horizontal and vertical directions as the co-ordinate axes, we see that the equations of motion arc as before.

FIG. 7.9


Let the particle hit the plane at A when t = t' , say. Then

Rcosa = ucosOt* (i) 1

R s i n a = usinOt* - -gt*2 2

(ii)

These are two equations in the two unknnowns R and t*. Substitute for t* from (i) into (ii)

R cos a R cos a 2

Rs ino = usin8 (-) u cos 8 - ig (-) u cos 8

- sin8cosa 1 cos2 a - R - -g cos 8 2 u2 cos2 8

R~

or

Therefore

sin8 cos a 1 cos2 a sin o =

cos 8 - Tg u2 cos2 8 R

2u2 cos2 8 - sin a cos 8 R = (Sin8coso g cos2 a

- 2u2 cos2 8 sin 8 cos a - sin a cos 8 - g cos2 a cos 8

That is

2u2 R=- cos 8 sin(8 - a ) .

g cos2 a

Time of flight t* is given by

Rcosa t =- ucos8

from (i)

- 2u2 cos a 2u sin(8 - a ) - cos 8 sin(8 - a ) . - - - - g cos2 a ucos8 g cos a

2u sin(8 - a ) t* =

g cos a

(iii)

To obtain maximum possible range, &,,, up the plane, we differentiate R with respect to 8 and set = 0.

dR 2u2 -- - [- sin 8 sin(8 - a ) + cos 8 cos(8 - a)] = 0 d8 g cos2 a


Thus the maximum range is obtained by pro- jecting in a direction bisecting the angle 5 - a! between the vertical and the inclined plane.

But cos A sin B = $[sin(A + B) - sin(A - B)]. Therefore

2u2 7r R,, = - . - [sin - - sin a]

gcos2a! 2 2 u2 1 - s i n 0 - - - - - u2 g 1 - sin2 a g ( l + sin a ) '

FIG. 7.10

Remark 7.2 To find t* without first finding R, either eliminate R using (i) or divide (ii) by (i) to obtain

Rsina! ~ s i n 0 t * - $ ~ t * ~ -= R cos a! ucos0t*

that is, 1 2 t a n a = tan0 - --t* t* = -(tan0 - tana!)cos6. 2 cos 0 9

This is the same as noting that on the plane y(t) = tana!x(t) for all t .

Remark 7.3 We could have used the product formula cosA sin B = a [ s i n ( ~ + B) - sin(A - B)] on R to obtain

2u2 R = - cos 0 sin(0 - a ) = - . 1 2u2 -[sin(20 - a ) - sin a].

g cosa g cos2 a 2

This clearly has maximum when sin(20 - a ) = 1 20 - a! = % as found by calculus.

Projectile problems involving inclined planes may also be solved using inclined axes in which the x-axis is taken along the plane and the y-axis perpendicular to the plane (see Figure 7.11).

FIG. 7.11

208 'Dynamics of a Particle

Relative to this co-ordinate system, the unit vertical normal B is given by

and

The equation of motion is r = -ge.

Taking components we have jE = -g sin a

Integrating twice and imposing the conditions i (0) = ucos(6' - a ) and x(0) = 0 gives

Similarly x = -gcosa

( 4

(vii)

(viii)

so that imposing the conditions y(0) = usin(6' - a) and y(0) = 0 we have upon integration

1 2 y = ut sin(6' - a ) - -gt cosa 2

(vii)

At A, y = 0 j ut sin(6' - a) - i g t 2 cos a = 0 and this implies

2u sin(6 - a ) t = O or

gcosa

2u sin(6' - a ) Hence time of flight-2' = as before.

g cos a

2usin(6 - a ) 1 2u sin(6' - a) Range = x(t*) = u cos(6' - a ) - -g sin a

g cos a 2 g cos a

- 2u2 2u2 sin2(6 - a) - cos(6' - a ) sin(6' - a ) - - sin a g cos a 9 cos2 a

- - - 2u2 sin(6' - a) [cos(6' - a ) cos a - sin a sin(6' - a)] g cos2 a

- - - 2u2 sin(6' - a) cos 6' 9 cos2 a

Example 8

A plane is inclined a t an angle of 4 5 O to the horizontal. A particle is projected from a point on the plane at ah angle tan-' to the line of its greatest slope. Show that the particle hits the plane a t right angles.

Solution

Y f

7.3 Projectile Motion 209 ' i

Suppose the velocity of projection makes an angle P with the plane. Then tan P = 3. Also the angle of projection 0 = 2 + /3 and a = q,

x = U C O S ~

x h=us inO-g t

At point of impact A on plane

221 sin(0 - a ) t = from (iv)

g cos a 2u sin ,d -- - g cos % ' 2

221 sin(0 - a ) 1! = from l iv\

-- - g cos % ' 2

Hence 7r ?r ?r

x(A) = u cos(- + p) = u{cos - cos P - sin - sin p) 4 4 4

and

7r 2u sin P 7r 2u sin P i ( A ) = s s i n ( - + P ) - g - = u ~ i n ( ~ + P ) - -

4 ( g - a ) cos 5

Hence v(A) is inclined t o the positive x-axis a t an y =- angle of -5. Therefore its inclination to the plane is - E - E - - -T . Thus the particle hits the plane

G W ~ a t right angles. ,

Alternat ive ly Using inclined axes. *=x

If the velocity of the particle a t A is perpendicular to the plane, then the velocity has no component in the direction of the plane. Thus x(A) = 0. But

x(t) = u cos(0 - a ) - gt sin a by (vi) 2u sin(8 - a )

= u cos(6' - a ) - g sin a g cos a

= u cos(0 - a ) - 2u sin(0 - a ) tan cr

210 Dynamics of s Particle

Therefore R

x(A) = u cos p - 2u sin P tan - for this problem. 4

Thus particle hits plane a t right angles.

Resisted vertical motion We now investigate the effect of air resistance on the vertical motion of a particle. From our experiences we can attest that air resistance is always in the direction opposed to the motion and its magnitude is a function, f , of the speed v . We consider here only the case where the resitance is proportional t o the speed so that f (v) = pv , say.

If we take the positive x-axis vertically upwards, then the fricitional force is given by -a2x. This is because for upward motion x > 0 and -a2x < 0 so that the resistance is downwards; also for downward motion x < 0 and -a2x > 0 giving an upward resistance. Thus -a2x always gives the correct direction for the air resistance.

The forces acting on a particle of mass m projected vertically upwards with intial speed u are gravity and air resistance. Hence its equation of motion is

where k is a positive constant. Thus dt - - 1 - -- dv g + k v '

Integrate with respect to v

k

Thus

That is

(ii)

(iii)

which upon integration yields

7.3 Projectile Motion 21 1

At t = 0, x = 0, therefore

1 1 0 = --(g + ku) + B B = p(g + ku) k

. Thus 1

x = ku) [1

At maximum height v = 0, which from (ii) implies

where t* is time of flight to maximum height. From (iv), the maximum height H,, is given by

From (ii) we have that as t -, oo,v(t) + - f . Thus as the particle falls from its maximum height (where v = 0) its downward speed tends towards f which is called its terminal speed. Note that this value is never quite attained since for large t ,

for all finite time. The particle returns to the point of projection when z = 0 that is

This is satisfied by t = 0 and is a transcendental equation for the positive value of t which also satisfies it. Note that if in any problem we require only a relationship between v and x, this is best derived by rewriting (i) as

which implies

This is integrable with respect to x. Now

. Thus

21.2 Dynamics of a Particle

and integrating with respect to x we have

At t = O , z = O , v = u hence

Therefore

which simplifies to

(vii)

In particular, at maximum height, v = 0 and x = Hmu Thus

as before.

Exercises 7.3

1 A particle is projected upwards in a direction inclined at 60° to the horizontal. Show that its speed when at its greatest height is half its inital speed.

2 A gun whose muzzle velocity is 60 m/sec is fired to strike a bird perched at a point a horizontal distance of 80 m and height 20 m from the mouth of the gun. At what angle to the horizontal must the gun be aimed? Take the acceleratibn due to gravity g as 10 m/sec2.

3 A particle is projected from a point at a height 3h above the horizontal ground, the direction of projection makes an angle a with the horizontal. Prove that if the greatest height attained by the projectile is h, the horizontal distance travelled by the particle before striking the ground is 6h cot a.

4 A mortar stands on a horizontal plane. It fires a bomb with velocity v a t a target on the plane. When the angle of elevation is a , the bomb falls a distance a short of the target. When the angle of elevation is ,d, the bomb falls a distance b beyond the target. ~ h 6 w th&t the angle of

a sin 2,d + b sin 2 a elevation required for the bomb to hit the target is

7.4 Simple Harmonic Motion

DEFINITION. Simple harmonic motion (SHM) is a linear motion of a particle in which the acceleration of the particle is always directed towards a fixed point and is proportional to the distance of the particle from the fixed point.

Equation of motion about centre of oscillation taken as the origin Take the origin at the fixed point and the line of motion as the x-axis. As shown in F@re 7.12 when the particle is a t P, its acceleration, being directed towardsdo, is in the direction PO. Similarly, when the particle is a t Q, its acceleration is in the direction QO while

7.4 Simple Harmonic Motion 213

its displacement is in the direction 03. Thus the acceleration is always in the direction opposite to the displacement, x . Hence -3 is parallel to x , so that -3 = w2x for some positive constant w Z where the square is used to emphasize the positiveness of the constant. Therefore,

This is the equation of simple harmonic motion.

acceleration *

FIG. 7.12

Equation (i) has three variables, namely, x , t and v . To be able to integrate it we need to rewrite it in terms of two variables only. Now

Hence (i) becomes dv 2

v - = -w x dx

(ii)

and integrating with respect to x , we have

where c is a constant of integration. Since v2 2 0 we must also have -;w2x2 + c 2 0 so that c 2 3w2x2 2 0. Actually unless x = 0 for all times, c must be positive. Let c = $a2w2 > 0 so that

(iii)

Note that a2 - x2 > 0 or 1x1 5 a so that the magnitude of the displacement of the particle does not exceed a . When the distance of the particle from 0 is actually a , its speed v = 0 so that the particle comes momentarily to rest at an extremity of its motion.

From (iii) dx dt 1 , v = - = dt

f w , / z F or -= * dx ~ 4 -

so that - f w d t . / (iv) "

The right hand side is immediately integrable and the left hand side may be integrated by making-.: the substitution x = a sin 0, say. This yields

dx a cos 0 dB

a2 - a2 sin2 0

214 ~ ~ n a m i c s of a Particle

Hence (iv) gives e = - f w t + a

where a is a constant of integration. Therefore

Since sin(a - wt) = sin[-(wt -a)] = - sin(wt - a ) and sin(wt - a + n) = sin(wt - a ) cos n + cos(wt - a ) sin n = - sin(wt - a) then

sin(a - wt) = sin(wt - a + ?r) = sin(wt + @).

Therefore no matter what sign is chosen for w in (iv) 1 i e solution is of the form

where /3 is a constant of integration. The motion of the particle is thus seen to be oscillatory in the interval -a. 5 x 5 a with

period 5 units of time. Note that since wt is an angle, w is measured in radianslunit time. The origin towards which the acceleration is always directed is called the centre of oscillation and the maximum displacement a the amplitude of the motion. The frequency, f , of oscillation is the number of oscillations the particle makes in unit time. Thus

1 W f = - = - or w = 2 n f .

period 2n

At the origin, the speed is maximum (= aw) and the acceleration is zero there. The solution (v) may be expanded to obtain

x = asinwtcos@+acoswtsin@ or x = Asinwt + Bcoswt ( 4

where A = a cos p, B = a sin /3. This clearly shows that two constants of the motion A and B or a and /3 must be determined to enable us obtain the motion in any specific problem.

Remark 7.4

An alternative method of integrating (i) is to multiply it by x to obtain 2 xx = -0 xx

which is immediately integrable with respect to t . This yields

as before. This method is called multiplication by an integrating factor.

Equation about origin different from centre of oscillation

If the origin of ceordinates is not a t the centre of oscillation the equation of simple harmonic motion (SHM) has a slightly different form. Let 0 be the origin and A the centre of oscillation (see Figure 7.13).

FIG. 7.13


Let x , s be the displacements of the particle at P from 0 and A respectively. If 1031 = k , then x = k + s so that

x = s and x = s.

The equation of SHM 2 s = - W s

becomes i -w2(x - k ) = -w2z + w2k.

As the motion is independent of the observer, this too is SHM. Thus the equation

2 = -w2x + constant

is also the equation of simple harmonic motion but with the centre of oscillation a t a point other than the origin.

Conversely if an oscillation of simple harmonic motion is given in the form

2 = -w2z + b, b a constant,

we can determine the position of the centre of oscillation by a translation of co-ordinates.

Let x = y + k where k is a constant to be determined. Then x = y, x = y so that simple harmonic motion equation becomes

ji = -w2(y + k ) + b = -w2y + b - w 2 k .

For motion about the centre of oscillation (y = 0), we must have

Hence the centre of oscillation is at x = k = 3. Mot ion of a loaded elastic springlstring

When forces are applied to the ends of an elastic string or spring, the stringlspring is stretched and is said to be in tension as it now has internal forces tending to counteract the external forces. According to Hooke's law the extension of the string, that is the increase in its length, is proportional to the applied external force. When the force is removed the string reverts to its original length, called its natural length. However, for each spring/string the above description is correct only for extensions within a certain limit, called the elastic l imit of the spring/string. Beyond its elastic limit small increases in the applied force produce disproportionately large increases in the extension of the springlstring. Furthermore, when the force is removed the string can no longer revert to its natural length.

Consider an elastic string/spring of natural length I. For an extension, x, within the elastic limit, let the applied forces induce a compensating tension T in the string. By Hooke's law

where k is a constant to be determined. Let A be the force which would double the length of the string thus giving an extension of I. Then


Hence X

T = -2. 1

(viii)

The constant X is called Young's modulus of elasticity.

Example 9

One end A of a light elastic string of natural length I and Young's modulus of elasticity, A, is fixed. To the other end is attached a particle of mass m and the assembly hangs freely under gravity. The particle is pulled down a further distance p and released, Show that initially the motion of the particle is simple harmonic with period 2 n m . Determine its centre of oscillation and describe the subsequent motion.

Solution

Let e be the extension of the string at EP when the particle hangs at rest in equilibrium (see Figure 7.14). Let x, measured downwards, be the extension of the particle beyond EP at time t and T the tension in the string. At EP, T-mg = 0 since the forces in the particle are a t equilibrium. 2 j D ~ u i l i ~ p s i t i m

X m g = - e

1 (*I When the particle is a distance x beyond E P ,

T = X(x + e l .

7 mg

1 FIG. 7.14

But the equation of motion of the particle is

where w2 = 5. This is the equation of simple harmonic motion with period % = 2r\/F and centre of oscillation a t the origin of cclordinates. Hence the equilibrium position is the centre of oscillation of the particle. But

v2 = w2(a2 - x2) from (iii).

Hence since v = 0 at x = p and t = 0,

Then

x = p sin(wt + a), say.


Therefore

7r 7r x = p sin(wt + -) = p sin(- - wt) = p cos wt.

2 2

But the motion is simple harmonic only as long as the string is taut and has a tension. The particle will where possible oscillate about EP with amplitude p. If p < e the particle does not reach as high as N L in its motion about EP. The string remains taut throughout and the motion is entirely simple harmonic. If p > e then when the phticle is at N L , it has a non-zero speed, there is no tension in the string and the equation of motion changes to y = -g where y is its distance above N L . It therefore moves like a particle projected vertically upwards to a maximum height where its velocity vanishes. It then falls freely under gravity until it reaches N L with the same speed as before but now in the downward direction. At this point the string becomes taut and the particle resumes simple harmonic motion about EP. If p = e , the particle just reaches N L as its maximum height (i = 0) at which point T = 0 but with the string taut. Although its velocity is zero, its acceleration however is maximum and equal to pw2 downwards. The particle then moves downwards due to this acceleration and T becomes non-zero. The entire motion is again simple harmonic although T becomes momentarily zero at the highest point of motion.

In summary: (i) If amplitude 5 e, the displacement produced by the mass, motion is entirely simple harmonic

with u2 = ; I . (ii) If amplitude > e , motion is partly simple harmonic and partly projectile.

In the latter case the motion is still periodic and the period = time for projectile motion from N L to maximum height and back + time for simple harmonic motion from N L to maximum depth and back.

Mot ion of a simple pendu lum Consider the following problem:

One end A of a light inelastic string of length I is fixed. To the other end is attached a particle of mass m which hangs freely under gravity. The particle is pulled sideways through some distance keeping the string taut and released. Show that for small initial displacements the motion is approximately simple harmonic with period 2 . 7 r m . Determine the centre of oscillation.

Solution

Take the fixed point, A as the origin of co-ordinates. At any time t let the particle be at the point P. The forces acting on the particle are gravity vertically downwards and the tension in the string in the direction PL. Thus the equation of motion of the particle is

Take the downward vertical AB through A

A as the positive x-axis (see Figure 7.15), the positive y-axis in the vertical plane containing A P and in the direction of increasing 6' where 6' is the angle that A P makes with the positive x-axis. Take the positive z-axis pointing out of the plane of the book to complete a right handed co-ordinate system. Thus

T = T(cos6i+sin8j) , g = g i ,

r(0) = I cos a i + I sin cr j, i(O) = 0. BY x Ag FIG. 7.15

The z-component of equation (i) is ~ ( t ) = 0. Integrating with respect t o i yietds i f f ) = q. Since r(0) = O,i(O) = 0 a cl = 0 and i ( t ) = 0. Integrating again we have z(t) = c2. But z(0) = 0 0 = cz. Hence z(t) = 0 for all t and the motion of the particle lies in the xy plane, i.e. the vertical plane containing the initial radial vector.

Now, let us use polar co-ordinates for the motion of the particle. Since Irl= 1 = constant, 1: = 0 and the radial and transverse components of the acceleration reduce to T 1 ~ 2 and l e respectively. Also the components of the velocity in these directions reduce to 0 and 16 respectively confirming that all motion is tangent to the string so that v = 16'. Resolving the equation (i) along the radial and transverse directions, we obtain

1 - 1 e 2 = g cos 6' - -T

m

and l e = -gain8

respectively. Using

equation (iii) becomes . de

16'- = -g sin 6' de

which is immediately in.tegrable with respect to 8 to obtain

1 x162 = g cos 0 + cs.

Initially 8 = a say, v = le = 0 3 6 = 0. Therefore

0 = gco8a+c3 - - r c g = -gcosa.

(iii)

Thus

or


Since ea 2 0, then 2g(cw 0 - cos a ) 2 0 * cos 0 - cos cr 2 0 or cos 8 2 cos cr * 181 5 cr. So the particle oscillates in the range -cr _< 8 5 cr. Also cos 8 2 cos cr =3

1 -T = g cos 8 + 1e2 2 g cos cr from (ii) and since 19' 2 0. m

Thus the string remains taut throughout the motion. To find the period of oscillation we solve for 9 from (iv)

so as to integrate with respect to 8. This cannot be done using elementary functions. However, if 0 is small (5 10' < 0.2 radians), we can obtain a good approximation, In that case sin8 m 8 and equation (iii) becomes

lI j% -ge

This is simple harmonic motion of period 27r . fi Applications of simple harmonic motion We give two areas of application of simple harmonic motion.

1 The simple pendulum is an instrument for measuring g, the gravitational acceleration. Since T = 2 n a , then T2 = 4n21Ig and a graph of T2 against 1 gives g.

2 Simple harmonic motion is used in construction timing devices in mechanical clocks and watches.

a. the simple pendulum is the timing device in so called grandfather clocks. b. mechanical clocks without pendulums have as timing devices oscillating balance wheels

with heavy rims to which hairsprings are coupled (see figure below). As the wheel rotates about a jewelled axis through its centre, the spring is coiled and produces a restoring couple proportional to the twist from the unstressed or zero position. Newton's second law in the form

' dv rn- = F

dt (i)

equates the rate of change of linear momentum to the applied force F. There is a corresponding equation governing rotational motion of a particle.

DEFINITION. The angular momentum of a particle is the moment of its linear momentum about the origin.

Take the vector product of (i) with r

(ii)

220 Dynaanics of a Particle

However,

Therefore (ii) may be written as

- (iii)

which shows that the rate of change of angular momentum equals the moment of the applied forces. When dealing with a collection of particles, the equation (iii) is added for all such particles.

A particle of mass m on the rim of a balance wheel of radius a has no radial component of velocity and a transverse component of a0. Hence its angular momentum is a(ma9)k = ma2dk where its motion is taken to lie on the xy plane. Every particle on the rim has the same instantaneous angular velocity. Therefore, adding for all particles we have

where M = total mass of wheel. For the wheel G = -pdk since restordhg couple is proportional to the twist. Therefore equation of motion of wheel is

Thusitsmotionissimple harmonic W

Remark 7.5

Because the rim of the balance wheel has a non-zero thickness and of the presence of spokes on the wheel, the summation of angular momenta for all particles of the wheel does not give exactly Ma2 but M12 for some constant 1 close to a. The sum C, mr2 is called the moment of inertia of the wheel about its axis of rotation.

Example 10 '

One end of a light'elastic spring of natural length 1.5 m and modulus 6g N is fixed. A particle of mass 2 kg is attached to the free end of the spring and held so that the spring is just fully stretched without tension and then let go.

a. Determine the amplitude and period of the motion.

b. When is the particle 0.75 m below its initial position?

Solution

We find the position of its centre of oscillation, EP,(see Fig. 7.16).


At EP,2g = +e = & e a e = t m . At an arbitrary displacement x below EP, .

Therefore x = -2gx

which is the equation of simple harmonic motion 27r

of period - FIG. 7.16 2s

45 set. Its general solution is -

x = ~ c o s f i t ++s in f i t

so that

) = - & ~ s i n & t + & + c o s f i t

Thus 1 1

- - = A . l + B . O * A = - - 2 2

and

Hence

Amplitude = lxmax 1 = m. The particle is 0.75 below its initial position NL if x = 0.75 - 0.5 = 0.25.

Therefore 0.25 = -i cos &t I coo f i t = -0.5 2

Example 11

An elastic string of natural length 51 and modulus A is cut into two parts of lengths 21 and 31. Each part is tied to the same particle of mass m. Their free ends are tied to two fixed points a distance 71 apart on a smooth horizontal table. Find the period of oscillation when the particle is displaced along the string.

222 Dynamarmcs of a Particle

Solution

FIG. 7.17

Let the two fixed points be A and B. See Fig. 7.17. Let the particle be at P, a distance x from A. The forces acting on P are TI along PA and Ta along PB. Let the string AP have natural length 21. Then the equation of motion of P is

Therefore

which is simple harmonic motion of period 27r E,

Exercises 7.4

1 Show by direct substitution that x = asin(wt + P) is a solution of the equation x = -w2x for arbitrary values of a and p.

2 Find the length of a simple pendulum if its period for small oscillations is 1 sec. Take the acceleration due to gravity g as 10 m/sec2.

3 If the length of a simple pendulum ie increased by 1% by what percentage is its period for small oscillations increased.

4 The length of a simple pendulum is 0.25 m and its maximum angle of swing is 8O. What is the frequency of the oscillation?

5 A body of mass 4 kg is hung on a light spring and is found to stretch it 3 cm. The mass is then pulled down a further 2 cm and released. Find the period of oscillation of the mass and its maxinium speed.

6 An elastic spring extends a distance I when a given mass is attached to its lowest point. The mass is pulled down a further distance a(< I ) and let go. Find the period of the simple harmonic motion that ensues and find its maximum speed.

7.5 Impulsive Motion 223

7.5 Impulsive Motion

Each one of us may attest from personal experience that if a running person bumps into an object- whether stationary or moving-the person feels as if he is given a blow by the object. At the moment of contact between them, the person experiences a change of velocity but no corresponding change in position or displacement. Indeed when two objects collide forces which could be very large but acting for an extremely short period of time are generated. Because these forces exist for such brief moments of time, it is difficult to describe or measure them directly: the only way to do so appears to be in terms of their effect on the velocities of the colliding objects.

In this section we learn how the mathematical (Newtonian) model of dynamics handles collisions.

DEFINITION, A large variable force I(t) which acts only for a brief internal of time t l 5 t 5 tz

is called an impubive force or blow. The time integral of I, It' I(t) dt, denoted by J is called its

impulse.

If an impulsive force is acting on a particle, its impulse

= mv(t2) - rnv(t1) = momentum immediately after impact - momentum immediately before impact

= change in momentum of particle due to impact

Effect of non-impulsive forces during impact Consider a particle of mass m under the action of impulsive forces I(t) in the interval t l 5 t 5 t2 = t l + A, A very small. Suppose it is also acted bn by non-impulsive force F. Since A is very small, F may be regarded as constant throughout the interval. The equation of motion of the particle is given by

so that

Thus

As A + 0, AF + 0 and its contribution to momentum changes is negligible. Thus we consider only impulsive forces during impact.

Conservation of linear momentum Consider two particles A and B of masses ml and m2 respectively. Suppose that they collide and that A is acted on by impulsive force I1 external to both particles whereas B is acted on by impulsive force I2 external to both particles. Let the impulsive reactions between the two particles be R o n A and therefore -R on B by Newton's third law. If their velocities at time t ate v l and va respectively then the equations of motion are


and

(ii)

Adding these equations we have

d ~ ( m i v t + m m ) = 11 + 5 = I say (iii)

where I is the total impulsive force external to the system of A and B. We note that the impulsive forces R , -R on A and B respectively are external when the

motions of A and B are considered separately. However when A and B are taken together as a system, these internal (i.e. to the new system of A and B) impulsive forces cancel out.

Suppose the component of I is zero in a fixed direction a. Take components of (iii) in the direction of a

Since

equation (iv) may be written as

On integration this gives

a . (mlvl + m2v2) = constant

This result can easily be extended to a system of many particles. Thus

the component of the total momenta of a system of particles is constant in any direction in which the components of the external impulsive forces acting on the system is zero.

This is the law of conservation of linear momentum during impact.

Collision of two smoo th spheres Consider the collision of two small smooth spheres A, B of equal radii and masses ml , m2 respectively. Let ul, u2 be the speeds of A, B respectively just before impact and vl., v2 their respective speeds immediately after impact. Let the corresponding velocities make angles al, a2 and PI, p2 with the -


line of centres as shown in Figure 7.18.

Before impact I After impact

FIG. 7.18

In general the velocities before impact are known and the velocities after impact are to be determined. The problem thus involves finding the values of four unknowns.

If both a1 and a 2 are zero so that the velocities of A and B before impact are along their line of centres, we have direct impact; otherwise we have oblique impact of the two spheres.

Whether it is a case of oblique or direct impact, the only impulsive forces acting on the spheres are produced by their reactions and act along their line of centres. Let these generate an impulse J ee in Figure 7.18a.

a. Motion perpendicular t o line of centres There is no impulsive force in this direction and linear momentum is conserved for A, B and A plus B together. Hence

mlu l sin a1 = mlvl sin PI for A

or ul sin a1 = v1 sin P1

or

and

m2u2 sin a 2 = m2v2 sin P2 for B

u2 sin a 2 = v2 sin pz (vii)

mlul s i na l + m2u2 sin a 2 = mlvl sin + m2v2 sin p2 (viii)

for A plus B together. Equations (vi) and (vii) show that the components of the velocities of the spheres are not changed in the direction perpendicular 'to their line of centres, along which the impact occurs. In particular for direct impact where a1 = a 2 = 0, the comporrents of velocities after impact perpedicular to the line of centres remain at zero, their initial values. Equation (viii) is equal to ml(vi) + m2(vii) and thus is not different from these. Therefore not more than two of them may be used at any time.

b. Motion along the line of centres There is no impulsive force in this direction for A plus B together. Therefore

Thus by considering conservation of momentum we have three equations, namely (ix) and any two of (vi), (vii), (viii). We need one more equation to enable us determine all the 4 unknown variables ul 1 ~ 1 , P1 and P2.


That equation is provided by an experimental law (credited again to Newton) relating the relative velocities with which the particles approach and separate from each other along the line of centres. The law states:

the component of the relative velocity of the particles along the line of centres immediately after impact equals (-e) x that immediately before impact.

where e is a constant with value in the closed interval [0, 11. In terms of Figure 7.18 this means

v2 cos P2 - v1 cos Pl = -e(u2 cos a2 - ul cos a1) ( 4

Note that for collision to occur ul cos a1 > u2 cos a 2 and there will be no separation unless v2 cos p2 > vl cospl. The constant, e, is called the coefficient of restitution or coeficient of elasticity. It depends only on the two colliding objects. If e = 0 the objects are inelastic and ua cosp2 - v l cos& = 0 so that there is no separation and the two objecta move with the same velocity. When e = 1 the objects are perfectly elastic. Finally, having found v l , v2, P1 and p2 we can now find the impulse J between the two particles

J = rn2[v2 ccs P2 - zs2 cos ail from B

O f

- J = ml[vl cos p1 - ul cos all from A

since change 4 = final 4 - initial 4 for any entity 4.

Example 12

A sphere of mass 40 gm is moving with constant velocity 20 m/sec. A second sphere of the same size but with mass of 60 gm is moving along the same line as the first sph~re and with velocity 30 m/sec towards it. If e = $ find the velocities of the two spheres after impact. Determine also the

. impulse generated between them.

Solution

Let the velocities of the spheres be u, v m/sec respectively parallel to. the 20 m/sec velocity as shown.

Before

,40 gm mgin After , ,

This is direct impact, momentum is conserved for the two spheree together. Therefore

40u + 60v = 40 -20 + 60(-30) =$2u + 3v = -50 (0

7.5 .Impu&ive Motion 227

From restitution 3 3

v - u = --((-30) - 2 0 ) ~ --(-50) a v - u = 30 5 5

(ii)

Solving equations (i) and (ii) simultaneously gives

v = 2 m/sec and u = -28 m/sec i.e. 28 m/sec in opposite direction.

60 6 J = -(2 - (-30)) = - 32 = 1.92. Newton - sec

1000 100

Example 13

Two spheres of masses 3 kg and 4 kg on a smooth horizontal tables are connected with a light inextensible string of length a. Initially the string is slack and the 4 kg mass is projected along the table with velocity 21 m/sec.

a. Find (1) the velocity with which the 3 kg mass begins to move

(ii) the impulse generated in the string when it suddenly becomes taut. b. Is there any change in the kinetic energy of the system?

Solution

a. While the string is slack the 4 kg mass moves in a straight line at constant speed 21 m/sec and the 3 kg mass remains at rest. At the instant when the string becomes taut, there is an impulsive tension along the string which jerks the 3 kg mass into motion.

The two masses now move with the same velocity (see the figure above). There is no motion perpendicular to this direction since there is no impulsive force in that direction. (i) Momentum is conserved for both spheres combined:

(ii) Impulse J = 3(v - 0) = 3 .12 = 36 Newton-sec. b. Kinetic energy before jerk = . 4 . 212 = 2 441 =.882 joules Kinetic energy afte'r jerk = 4 - 7 . 1 2 ~ = 504 joules Hence there is a loss of kinetic energy of 882 - 504 = 278 joules.

Example 14

An object of mass 6 kg moving with velocity 8 m/sec collides with another object of mass 9 kg travelling with velocity 10 m/sec in a direction making an angle 60° with the velocity of the first object. If the two objects coalesce on impact, find the velocity of the combined mass.


Solution

8 - JW

a 6 kg 9 kt3

Before

Let u, v m/sec be the components of the velocity of the combined mass parallel and perpendicular to the 8 m/sec velocity (see figure above). Momentum is conserved for the system of two objects together. Therefore

1 15u = 6 .8 + 9 .lo cos 60° = 48 + 9 . lo(-) = 93.

2

Therefore 93

u = - = 6.2 m/sec 15

Also

Example 15

A ball of mass 5 kg moving with speed 30 m/sec impinges on a second ball of equal radius but with mass 15 kg moving at a speed of 10fi m/sec. If their velocities immediately before impact are inclined at angles 30°,450 respectively to the line of centres at the instant of impact, their velocities immediately after impact given that e = 4 . Solution

determine

Let the velocity of 5kg mass A be u m/sec in direction IY to line of centres and let the velocity of 15kg mass B be v m/sec in direction P to line of centres. Momentum perpendicular to line of centres:


Momentum along line of centres:

5(u cos a ) + 15(v cosp) = 5(30 cos30°) + 15(10ficos 45")

i.e.

or

Restitution:

i.e.

Adding (iii) and (iv)

1 v cos p - u cos a = --(lo& cos 45" - 30 cos 30")

2

(iii)

Dividing (v) by (ii) gives

? 4 + 2 5 1 4 c o t p =

1 10

= -(9&+ 10) cot p = -(9&+ 10). 4 16

Squaring (ii) and (v) and adding

Similarly using (i), (iii) and (iv) we solve for u and a.

Alternatively It may be more convenient to use components of unknown velocities parallel and perpendicular to the line of centres.

After

For example, if we use ul , ua and vl, v2 (see figure above) the equations above become (replacing u cos a , v sin p by ul , v2 respectively)

u2 = 15 from (i) va = 10 from (ii)

UI +3vl = 15&+10

( 4 (iia)

(iiia)

(iva)


Adding (iiia) and (iva) yields

and

Collision w i t h fixed object Consider the collision of a particle of mass rn with a fixed smooth object. The object thus cannot move as a result of the impact. Hence its velocity both before and after the impact is zero. Let the initial velocity of the impacting particles be u at an angle a to the line of impact and its final velocity v at an angle 0 (see Figure 7.19).

Before After

FIG. 7.19

There are now only two unknowns v, 0 and hence we need just two equations involving these to be able to solve for them.

Conservation of momentum: , . There is no impulsive force perpendicular to the line of impact. Hence

musine = musina =$ usin0 = us ina

Restitution: For motion along the line of impact

where positiveG is taken in the direction away from the fixed object. Thus

v cos 0 = ezr cos a.

Solve for v and 0. By division t . 1

tan0 = - tancr. e

(ii)

Squaring and adding v2 = u2(sin2 a + e2 cos2 a)

= ua cw2 a(tan2 a + ea)

or

Also

v = u cos a J-oc'. I ' 9 . [ l '

the impulse generated = m[v cos 0 - (-u cos a)]

= m[eu cos a + u cos a] from (ii)

= mu(1 + e) cos a.

The fixed object also experiences this impulsive force. If a steady stream of particles impact on the fixed object then the latter experiences a force equal to the rate of change of momentum = e.

Thus if water from a hose impinges on an object at a constant speed of u m/sec, the force on the object = mu(1 + e) cos a Newtons.

. - Example 16 I .

Find the angle at which a smooth sphere must be projected towards a fixed'horizontal plane so that the sphere is deviated through 90'. Take e = a. . T:

, \ \ \

c . \

Before After

Let the velcoity of the sphere before impact be u at an angle 0 to the normal to the plank [see the figure above). If it is turned through 90° then its velocity v after impact is inclined at $ - 0 to the normal. Momentum parallel to plane is conserved ' '

Restitution: 7T

vcos(- - 8 ) - 0 = -e(-ucos0-0) *vsin9= eucos0. 2

(ii)

Eliminate v between (i) and (ii), for example by division . .

3 d3 tanO=ecotB a t a n 2 0 = e = - 3 t a n $ = - A

2 0 = tan-' -

4 2 '

Loas of kinetic energy at impact ., ,

Since there is no change in the components of the velocity for either particle perpendicular to the line of centres/impact, the changes in kinetic energy (KE), if any, must be produced by the components of velocities along the line of impact.


The KE before and after impact due to the motion along the line of impact are

respectively.

"1 Before

'"1 "2 After

for any ml , m2, 41, 42 , we have

change of KE = final KE - initial KE

3 u t mlul coe a1 +m2u2 cos a 2 = mlvl cosP1 +m2v2 cosP2 by conservation of total momentum along line of impact. A h by reetitution

v1 cosP1 - v2cosP2 = -e(ul c o s q - u2coaa2).

Substituting into (ii) for v's we have

m1m2 [e2(u1 caa a1 - u2 coa all2 - (ul coa a1 - u2 cos a2)'] change of KE = - 2 m l + m 2

Thie is negative eince e < 1. Hence there is a loss in KE of amount

Applicatione The moet obvious application of collisions is in the game of billiards. Players must hit a ball lying on a smooth table to impart to it the desired velocity to enable it collide with another stationary bdl. After the impact the latter ball is required to move in a desired direction.

Another, application is in the manufacture of artillery gum. Before firing, a gun and the shell (bullet) within it are both at rest. During firing the eyetem of gun and ehell together have no

7.5 Impulsive Motion. 233

external impulsive force acting on it in the horizontal direction. Thus their total momentum in that direction remains constant at its initial value which is zero. However, since the shell on its own has a forward momentum, the gun must have an equal backward momentum. Thus a mechanism (usually springs) must be provided to bring the gun to rest and ready for the next firing.

Our final example is a weapon much beloved by the police. A stream of water under high preseure when trained on a protester will knock the person down. The change in momentum of the water impacting on the person produces a force or blow which has been known on occassion to knock someone senseless.

Exercises 7.5

Two eleastic spheres each of mass m collide directly. If the velocity of one of them is exactly reversed, show that the spheres were travelling towards each other. A smooth sphere A of mass 4 kg is tied to a fixed point 0 on a smooth horizontal table by a light inextensible string of length a. The sphere A is at rest on the table with the string taut. A second sphere B of the same size but mass 5 kg hits A directly with velocity 10 m/sec so that their line of centres at impact makes an angle of 30" with OA, the string remaining taut throughout. If e = i, find the velocity with which A begins to move. Hint: Why is the component of velocity of A zero along OA? Two particles A and B of masses ml and mp respectively are connected by a light inextensible string. The particle A moves on a smooth horizontal table and B hangs under gravity with the connecting string perpendicular to the edge of the table. Just before A reaches the edge of the table, the common speed of the particles is zc m.sec-l. Find the velocities of the particles immediately A leaves the table. A child is playing with two identical tennis balls, A and B say, on a smooth horizontal floor. The child places the balls on the floor with B between A and a smooth vertical wall in such a way that the vertical plane containing A B is normal to the wall. The child strikes A to impinge directly on B . The coefficient of restitution between the two ball is $ and that between a ball and the wall is ;. If A is given an initial velocity u , determine the velocity of A after its second impact with the ball B. The edges of a smooth horizontal table are raised. A small spherical ball is projected with velocity v to impinge on two adjacent edges of the table. Show that iLs velocity after the second impact is -ev where e is the coefficient of restitution between the table and the ball. A ball impinges obliquely on another ball a t rest. (i) Find e if the velocities of the balls immediately after impact are at right angles. (ii) What can you deduce about the relative masses of the stationary and moving balls? A particle moving with a speed of u m/sec in a direction making an angle of O0 with the horizontal strikes a smooth horizontal plane and rebounds. Find the velocity of the particle immediately after impact if the coefficient of restitution is e.

A smooth ball of mass m moving at 8 m/sec is striken by a second smooth ball of equal radius with mass 2m moving at 10 m/sec. If their velocities immediately before impact are inclined at 4 5 O and 30' respectively to the line of. centres at the instant of impact, determine their velocities immediately after impact if (i) the balls are assumed perfectly elastic (ii) the coefficient of reetitution between the is 3 .


7.6 Motion of Particles with Variable Mass

Recall Newton'e second law of motion (see Section 7.1), namely

the rate of change of momentum of an object is proportional to the external impressed force F acting on it.

In equation form this is

where the constant of proportionality has been made unity by an appropriate choice of the unit of force. For an object of constant mass m this reduces to the familiar equation

that we have used so far in this chapter. ~owe"er, there are systems of practical importance where the mass of the particle/object is not constant. Some such objects are rockets, electrically charged par&les moving a t high speeds, and motor cars. In the car, like the rocket, as fuel is burned the mass of the object decreases. However for the car, unlike the rocket, the mass of the fuel is small ixi comparison with the rest of the object and so the mass of the car is regarded as approximately constant. The mass, m, of a charged particle is given by

where mo is a constant, v the speed of the particle and c the velocity of light. At small speeds m R rno but this no longer holds a t high speeds.

We briefly discuss the motion of a rocket.

Motion of a rocket The rocket was developed in response to military needs. We saw from the kinematics of a particle that the maximum range of a projectile on the horizontal ground is u2/g where u is the speed of projection of the particle and g is the acceleration due to gravity. However, an explosive mechanism is needed for imparting the initial speed u. Since there is a limit to the intensity of such devices, there is a corresponding limit on the speed that can be achieved and consequently on the maximum range.

The rocket operates by providing a propelling force'to the particle over a period of time. Thus much higher terminal speeds can be achieved before the particle moves freely as a projectile.

Consider a rocket which burns fuel (part of its mass) a t a constant rate of r kg/sec for t* seconds. During this period products of combustion are ejected from it a t a constant velocity c m/sec backwards relative to the velocity of the rocket. After t* seconds the rocket travels freely as a projectile.

Derivation of equation of motion I

Suppose external forces equal to F ( t ) are acting on the rocket at time t . In a small interval of time, the change of momentum of the rocket equals the impulse on the rocket. Let the velocity of the particle at time t be v. In a small interval of time, At, the velocity of the rocket increases to v + Av and the mags m breaks into two parts m + Am, -Am with velocities v6, ( v - c)6 where 6 is a unit vector in the direction of motion of the rocket. See Figure below.

7.6 Motion of Particles with Variable Mass 235

The impulse-momentum equation is

( m + A m ) ( v + Av)& + ( -Am) (v - c)e - mu& ' .

= I+At F( t ) dt (*I

Note that since the mass is decreasing A m < 0 and1 vG!fl n t + h - " . thus ( - A m ) is positive. However, the standard expression m + A m must be used. Expanding the , .

above we obtain, for F = Fe

m A v + ( A m ) v + ( A m ) A v - Am(v - c)

= I+At F ( t ) dt

Divide by A t , A m

At At

Taking the limit as At 0 dv dm dm

m- + -v - - (v - c) = F( t ) dt dt dt

( ii)

d dm z ( m v ) = F ( t ) + -(v - c) dt

. dm = F ( t ) - r(v - c) since - = -r

dt

1.e. d -(mu) = F( t ) + r(c - v). dt

(iii)

By comparison with the basic Newton's equation of motion (i) we see that the reaction of the ejected gases produces an impulsive force called thrust of amount r(c - v ) on the rocket in the direction of motion.

On simplification (ii) becomes

where mo

m = m ( t ) = mo - ~t for 0 < t 5 t* < - r

where mo is a constant, the total mass of full fuel and payload of rocket.

Remark

There is another interpretation of the left hand side of (*). Suppose that during the time interval [t , t + At] a particle of initial mass m travelling with velocity vZ picks up additional mass A m a t a constant rate and that the additional mass has velocity u6 = ( v - c)6. Let the combined velocity of the m + A m particle at time t + At be denoted by (v + A v ) ~ . Then the change in momentum

236 Dyn-CB of a Particle

d i c h is exactly the expression in left hand side of (*). Thus Am > 0 3 increase in mass as when a particle moves through a cloud of water vapour

which condenses on it. Am < 0 decrease in mass as in rocket motion.

Solution of Equation of Motion We consider the solution of equation (iv) for some specified values of F(t).

Case I: F(t) = -mg gmviiy with no resistance Equation (iv) becomes

dv dm m- = rc - mg using - = -r from(v)

dt dt = rc - g(m0 - rt)

dv m- .= rc - mog + rgt

dt

Since m > 0, the particle will rise immediately (at t = 0) if rc - mog > 0. It will never rise unless rc - mog + rgt* > 0. Now to integrate (vi):

dv dm dv c dm m-+c- = -mg* -+-- - - -9.

dt dt dt m dt

This last equation is integrable with respect to t. However in problems where 9 is constant, less manipulative complexity is encountered when m is used as the independent variable.

dv dv dm dv Thus - = - . - = -r- and (vi) becomes dt dm dt dm

Integrating with respect to rn we have

When t = 0, v = 0 and m = mo. Therefore

9 9 0 + clnmo = -mo + const. ,+ const. = c h m o - -mo. r r

Hence 9 9 v + clnm = -m+ c h m o - -mo r r

(viii)

dx dx dm dx Again using v = - = - - = -r- this becomes dt dm dt dm

(vii)


which on integration with respect to m becomes

But 1

/ ln (g) dm = m in (E) - / m(-) dm integrating by parts m

Therefore

-rx + cm o n (g) When t = 0, x = 0, m = mo and therefore

Thus - - rx + cm (In (g) - 1) = l ( m 2r - ma)' - cmo

At time t* when all the fuel has been burnt, let v = v*, x = x* and m = m*. If we set the mass of fuel when full to be Em0 where 0 < E < 1, then m* = (1 - E)mo so that

v* + cln (2) = 4(m* - mo) from (viii)

Note that - i).l( 0 since 0 < 1 - E < 1 SO that the first term on the right hand side of (x) is positive as is x' itself (see Problem 1 in the Exercises). Also from (ix)

m* 9 rx* = cm* In ( rno) + c(mo - m*) - -(m* - mo)l

2r

Therefore

or


See also Problem 1 of the Exercises for verification that z* > 0. The particle now moves as a projectile with initial speed v* from height x*.

Case II: F = -mg - kv gravity plus dmg/air resistance which is proportional to speed Equation (iv) becomes

dv dv dm Using - = - . - dv = (-r)- this becomes dt dm dt dm

(xii)

This may be integrated by multiplying by an integrating factor m-kl r , the derivation of which is beyond the scope of this course. Thus

This last equation is immediately integrable to obtain

so that cr

v = - m + - + A m k / ' . r - k k

dx dx dm dx Setting v = - = - . - = -r- in (xiii) and integrating yields

dt dm dt dm

as the equation for x.

Exercises 7.6

(xiii)

(xiv)

x2 x3 xP 1 By using the expansion ln(1- x) = -x - - - - - . . . - - - . . . show that v* and x* are

2 3 , r both positive.


dv c dm 2 Starting from - + -- = -g integrate with respect t o t and apply the initial conditions

dt m dt v = 0, na = mo a t t = 0 t o obtain

v + c ln (mo - r t ) = -gt + c ln(mo).

Writing v = x, integrate this again with respect t o t to obtain

Verify that this is consistent with equation (ix) of the text,


1 A particle is projected vertically upward with a velocity of u m/sec and after T sec another particle is projected upwards from the same spot with the same velocity. Find where the two particles will collide and the time at which the collision occurs.

2 A gun mounted on a cliff of height h m above sea level shoots a cannon with muzzle speed U . Determine the farthest horizontal range of the gun measured at sea level and the angle of projection t o achieve this range. Hint: Find point on sea which can just be reached.

3 A light inextensible rope passes over a fixed pulley. At one end of the this rope is a body of mass MI . At the other end of the rope there is a pulley of mass M2 over which passes another light inextensible rope with bodies of masses nal and m2 attached a t the ends. Prove that the acceleration of the mass ml is

4 A gun fires two shots, the nozzle speed in each case being u. In the first case the shot is fired at an angle of elevation a , and in the second case it is fired a t an angle of elevation P, with a > p. Show that the time interval between the two firings such that the two shots will collide

2u sin(a - p ) in midair is -

g c o s a + cosp, 5 Two smooth spheres of mass m each moving with velocity U in directions a t right angles to

one another, collide in such a way that the line joining their centres is the direction of motion of one of them. Find the velocities of the spheres after impact if e = $.

6 A ball is dropped on the floor from a height of h. If the coefficient of restitution between the ball and the floor is el find the height of the ball a t the top of its nth rebound.

Answers to Exercises Chapter 1

Exercises 1.1 (Page 7) 1 6

5 ST2 = T R ~ + R S ~ i.e. 40 = 8 + 32 =+ LR is a right angle.

7 (9 ,5)

Exercises 1.2 (Page 11) 1 324 sq. units

2 Area is zero

3 (i) S(21,13) (ii) area = 165 sq. units.

4 C(-14,20); angle B is a right angle; area = 369 sq. units.

5 lPQl = lSRl = m, fQRI = IPS] = m; area = 10 sq. units.

6 Area = - i ( t l - t2)(t2 - t3)(t3 - t l ) # 0 if t1,t2,t3 are distinct.

7 Area = (tl - t2)(1 +t1t2) = 0 if 1 +t1t2 = 0 or t l = t 2 .

8 Area = 423 sq. units

10 Area = 1343 sq. units.

11 Area = 554 sq. units.

Exercises 1.3 (Page 24) 1 (i) 3 x + y = 1

(ii) 3y - 72 - 5 = 0 (iii) (tl + t2)y - 22 = 2at1t2 (iv) a(cos e 2 - cos 0l)y - b(sin e2 - sin O1)x = ab sin(O1 - 82) or

a s i n ~ y + b c o s ~ z = a b c o s ~ .

2 x - 2 y = 4

3 x Y + - = 1 intercepts are -;, -5 on 2- and y-axis respectively -7/3 -714 2 . (ii) - Y + - = 1 intercepts are -$, c on x- and y-axis respectively

-c/m c x

(iii) - Y + - = 1 intercepts are -f , -$ on x- and y-axis respectively. -n / l -n/t

1 4 y + x x = 5 13 5 Slope = +, y-intercept = 7

Answers to Exercises 241

6 x + y = l

8 2 = &J-. All points of the curve are the same distance from the fixed point (g, f).

Exercises 1.4 (Page 42) arctan 7

(i) LABC=90° (ii) L B AC = 45O

(i) 8 y - x = O (iii) 1 5 2 + 2 1 y - 4 7 = 0

(ii) 52 - y - 13 = 0 (iv) 3 y - 3 x + 7 = 0

Rewrite equation as 2x+ y - 5+ t (x+ y - 7) = 0; fixed point is (-2,9) the point of intersection of the two lines 2x+ y - 5 = 0 and x + y - 7 = 0. k = 2 o r -4

1202 - 20y + 11 = 0 is perpendicular to x + 6y - 7 = 0; 40% + 60y - 73 = 0 is perpendicular to 32 - 2y + 2 = 0.

(i) 22 - y = 0 and 22 + y = 0 (ii) 22 + y = 0 and x - 3y = 0 (iii) y - x + 2 = O a n d y + x + l = O (iv) 3 ~ - 2 y + 3 = O a n d x + y - 7 = 0

(i) xy - 4y2 = 0 (iii) y2 - (m + n)x + nmx2

(ii) 3x2 - 8xy - 3Y2 = 0

arctan 9 45O

tan 8 = in each case (8 = angle between lines).

Miscellaneous Exercises (Page 43)

4 (k2 - 1)x2 + (k2 - 1 ) ~ ~ + 2(k2 + 1)x - 2(k2 - 2)y + 2k2 - 5 = 0; straight line; 42 + 2y - 3 = 0. 1

5 -- 4 , 1 ,6 6 x 2 - 6 y + 9 = O

11 A = tan-' (I;i?) = 57.53O, B = tan-' (y) = 74.74O, 6 = tan-' ( G ) = 47.73O

12 (i) 3y = 122 + 4 (ii) 4 y = 3 x + 1 (iii) 12y = 33: + 1 (iv) a Y = 3 x + 1

242 Answers to Exercises

14 (i) x = 2; y = -3 . ( i i ) x - y + 2 = 0 ; x + y + 2 = 0 . (iii) 2 2 - y + 1 = O ; x + 2 y - 1 = 0

3 15 tan-' - = 36.87'

4

16 Point of intersection = ( -2 , l ) ; angle between lines = 60'.

18 2y2 - 7x9 - 2x2 + lox - l5y + 20 = 0.

19 Equation reduces to (y + x sin a ) 2 + (x cos a + a)2 = 0. This gives a pair of lines y + x sin cr = 0, x c o s a + a = 0 except when cr = *;.

Chapter 2

Exercises 2.1 (Page 47) 1 Centre (i, 1)) radius = +I/@

(ii) Centre (2,3), radius = 2

(iii) Centre (1, - g), radius = $m (iv) Centre (-1, !), radius = fi (v) Centre (1,2), radius = 6.

2 (i) No; imaginary radius. (ii) No; coefficients of x2 and y2 are not equal. (iii) (x - - ( y + 3)2 = 2. No; negative sign between squares.

1 1 17 (iv) (x - + (y + -I2 - - +

2 - , Yes.

Exercises 2.2 (Page 50) 1 5x2 + 5y2 - 11x - 9y = 12

2 y = - 2 x

3 (11-6) 4 2x2 + 2y2 - 412 + 7 = 0

5 x 2 + y 2 - 6 x + 4 y = O

6 x ~ + ~ ~ + ~ x + ~ ~ + c = O , centre (-1,-l),radius

7 x 2 + y a - 3 2 : + 2 y = O

8 x 2 + y 2 - 4 x - l O y + 1 6 = 0

9 4x2 + 4y2 + 212 - 27y - 14 = 0

10 ( x - x ~ ) ( x - ~ ~ ) + ( ~ - Y ~ ) ( Y - Y ~ ) = o o r x 2 + y 2 - x ( x 1 + x 2 ) - ~ ( ~ 1 + ~ 2 ) + x 1 ~ 2 + ~ 1 ~ 2 = 0.

Exercises 2.3 (Page 63)


7 2 units

8 units

9 x + 2 y + 4 = 0 , y + 2 x - 1 = 0

10 Their gradients are 2, -+ 11 3 2 - 4 y + 7 = 0 , 4 ~ + 3 y - 1 = 0

12 y = f i

13 x - 3 y + 1 7 = 0

14 x + y + 1 = 0 chord of contact; tangenh y - 2x = 5 , x - 2y = 5

15 2 ~ + 3 y - 3 = 0

16 From origin: hx + k y + d = 0; from ( h , k ) : 2hx + 2ky + h2 + k2 + d = 0

17 ( 0 , -1) and ( 3 , -2)

Exercises 2.4 (Page 72) 1 x2 + y2 - 2ax - 2by = 0

Miscellaneous Exercises (Page 75) 1 k = 2 4

2 ~ ~ + ~ ~ - 4 x - 4 y - 9 = 0

3 7 x 2 + 7y2 - 61x - 25y + 52 = 0

5 ~ ~ + ~ ~ - 2 ~ - 1 = 0 a n d x ~ + ~ ~ + 6 ~ + 7 = O

6 b = 5; lines intersect at ( 3 , l ) ; tangent to given circle is 4y - 3x + 5 = 0.

7 ~ ~ + ~ ~ - 2 x - 3 y = O (i) circles through vertices of a traingle taken in pairs. Circumcircle is common to all 3

systems. (ii) take four distinct points on a circle: there are six coaxial systems passing through pairs

of the four points. The original circle is common to all the six systems.

9 A = -5, p = 5. Circumcircle is 3 ~ ~ + 3 ~ ~ - 6 3 : - 8 y = 0 . Centre of orthogonal circle is ( 1 , -!), with equation 2x2 + 2y2 - 42 + 3y = 0.

1 1 10 y = ( 1 + - a ) z and y = ( 1 - - m ) x or 5x2 - 24xy + 12y2 = 0 as a line pair. Angle between

6 6 them = arctan &a.

244 Answers to Exercises a2 16 (a+X)x+by+h+e=O; p* =(-a,+), b#0

17 (12+m2)x2+(12+m2)y2+2(hm2-mk~l+n1)2+2(kl2+nm-lmh)y+d(l2+2)+2(n-lh-k)n = 0

18 ax2 + ay2 + 2(ha - r21)x + 2(ka - r2m)y + cda - 2r2n = 0 where r2 = h2 + ka - d and a=hl+km-n.

Chapter 3 Exercises 3.2.1 (Page 80) 1 16x2 + 19y2 + 4xy - 42 + 2y - 1 = 0 2 x2 + y" 22y - 122 - 12y - 12 = 0 3 11x2 - 96xy + 39y2.+ 1202 - 210y + 75 = 0 4 9x2 + 5y2 - 182 - 4y - 19 = 0 5 z2 - 42 - 2y + 7 = 0 or x2 - 4x - 18y + 103 = 0 6 25x2 + 16y2 - 2002 - 96y + 144 = 0 or 25x2 + 16y2 - 2002 - 296y + 1744 = 0 7 (1 - e2)z2 + y2 - 2~3: + c2 = 0.

Exercises 3.2.2 (Page 84)

1 (2 + f12 + (Y - $I2 = I ellipse: centre (-i, %) 1/2 3/4

(I/ + ;I2 - (x + %)2 1/36

= 1 hyperbola: centre (-; , - B) 1 /48

3 (2 - 4j2 (Y - I)= -- 4 10 - 1 ellipee: centre (+,I)

(y - 4)2 (2 + +)2 4 ----- - 1 3

= 1 hyperbola: centre (-4,4)

5 (y - i)2 = -3(z + 1) parabola: vertex (- 1, i) 6 (3: + 2)2 = $(y - f) parabola: vertex (-2, f) 7 5(x + 2)?+ 6(y - $)2 = 0 single point: (-2, i) 8 10(z + 2)2 + 12(y - i)2 = -30 empty set.

10 y = 2a(x - a). Exercises 3.3 (Page 97)

Equations of directrices y=f3&


2 Equslions of directtices: X = f -$J Equations of aeymptotes: Y = &fix

Equations of Equations of

Equations of directrices: X = f 9

asymptotes are Y = &AX k*, Y directrices X = f

Equation of directrix is Y = - 2

246 Aoswers- to Exercises

Y - " 6 Equation of directrix is X = -2

Equation of directrix is X = 1

Equations of asymptotes are Y = f +X Equations of directrices Y = &-&

Equations asymptotes: Y directrices Y =

Equations of asymptotes are X 2 0, Y = 0 Equations of directrices X + Y = f 1

Answers to Exercise* 247

Equations Equations

asymptotes: X = 0, Y = 0 directrices Y - X = f 2

Vertex (0, $) Equation of directrix Y = 3 s y = 2

Rewrite as (Z + +)= (g - 1 2

- 1 +-- Centre (- $, 1) Equations of dire~trices Y = f 2 y = 3 or - 1


14 Centre C(2, -3) Y , k' Asymptotes: X = 0 x = 2 and -%, *. Y = O z y = - 3 Equations of directrices: ( = f = - y = r x + y = - I f .\/Z where X = ( + q, Y = ( - 17

15 '\ '. *'.

*. Centre (-! , 4) *% '. Equations of asymptotes: X = 0 z = -3 . 2

% a n d ~ = 0 ~ ~ = 3 \

Equations of directrices: [ = f & -'. c -'8 0 + x

\ >. *. = ~ = h t J z ~ . + r = - l ~ ~ -'. * X - 3+Ji I-& '\ WT,-+ %\ \'*. '. '.

'\ *. %,, '\

\'. '. 16 Centre(-1,-1)

Equations of asymptotes: Y = &2X ~ y = 2 x + l o r y + 2 ~ + 3 = 0 Equations of directrices: Y = f $6 ='y=-l+EJ 5 o r - 1 - 6 6

Y Y

f / 3 s f i 17 4(T ,T-4 ) (2 - 3)2 + (Y + 412

25/4 25 - 1 Rewrite ds - - -

Centre C(; , r 4 ) Equations of directrices: Y = ~tqfi - = y = - 4 f $4.

Exercises 3.4.1 (Page 100) 1 Tangent: y + x = 0; Normal: y - x = 0


2 Tangent: 4 y + x = 8; Normal: y - 42 + 15 = 0

3 Tangent: 16y + 3 s = 6; Normal: 9y - 482 + 361 = 0

4 Tangent: y = 1; Normal: x = 2

5 Tangent: 255x + 189y = 587; Normal: 1701x - 2295y = 2467

6 Tangent: x = 1; Normal: y = 3

d y 62 + 3 8 - = - - a Tangent: x = : ( - I + a); Normal: y = $

dx 6 - 4 y '

dy 6 - 6 z 10 -=-. a Tangent: x = Q; Normal y = -g d x 5 + 6 y 1

dy 11 - = - 6y + 10 ; Tangent: 7 y + x = 9; Normal y - 7 2 + 13 = 0. d~ 2y(4x - y + - 6 2 - 4

Exercises 3.4.2 (Page 104) 1 Tangent: 4y - x + 8 = 0; Normal: y + 42 - 15 = 0

2 Tangent: 32 + 16y - 6 = 0; Normal: 482 - 9y - 361 = 0

3 Tangent: y - x + 2 = 0 ; Normal: y + x - 6 = 0

4 Tangent: 39y + 532 - 147 = 0; Normal: 6892 - 637y + 2123 = 0 5 Tangent: 42 + 3y + 16 = 0; Normal: 122 - 16y + 123 = 0 6 Tangent: y = 6 ; Normal: x = -1

7 Tangent: 2 f i y - x + 8 = 0; Normal: y + 2 4 x = 3 8

8 Tangent: f iy - z = 1; Normal: y + f i x + 4 4 = 0.

Exercises 3.4.3 (Page 108) 1 a. y - 52 = 5 and y + 52 = 5 b. (2 , -5) and (-2, -5) respectively.

2 a. 4y - x + 8 = 0 and x = 4 b. ( 2 , -;) and (4,O) respectively.

3 a. 2y = z + 3 and y + x = 0 b. ( 5 , 4 ) and ( 2 , -2) respectively.

4 No tangents as equation for m, 123m2 - 244m + 128 = 0 has no real roots. Point ( - a , 1) is on same side of hyperbola as focus.

5 No tangents. as equation 3m2 + 2 m + 8 = 8 has no real roots. Point ( 1 , l ) is inside ellipse.

6 a. 8 y + 2 x - l = O a n d y + x = 2 b. (1, -$) and ( 4 , -2) respectively.

7 No tangents as m2 - m + 2 = 0 has no real roots. Point ( 1 , l ) is on same side of parabola as its focus.

8 a. 2y - 5x + 20 = 0 and 18y - 52 - 60 = 0 b. ( 2 , -5) and (-6, g ) respectively.

9 c = O , , m # 0 (Tangent not possible if m = 0) .

250 Answfirs to Exercises

Exercises 3.5.1 (Page 112) 1 z = 2cos0, y = 3sinO 2 z =2t2 , y =4t

Exercises 3.5.2 (Page 116) 14 26 1 (i) (g, 6) and (-2, -2) (ii) (0,4) and (g, T) (iii) (0.4) and (-+,0).

3 Tangent: 5 4 + 42 = 52 + 1 0 d ; Normal: 4y - 5 6 x = 8 - Yf i 4 Tangent: & + 22 = 18; Normal: 2y - f i x = 5fi 5 Tangent: y = &x + 3 - 6 6 ; Normal: fi + x = 12 + 3& 6 Tangent: y = 32 + 2; Normal: 3y + x + 4 = 0

7 Tangent: 4y - 22 + 1 = 0; Normal: 4y + 8z = 9

8 Tangent: y = x + (2 - %)a; Normal: y + z = $a

9 Tangent: y + (fi - 1)z = b; Normal: (A - l)y - z + 3( f i - l)b = 0

10 Tangent: y + (2 + &)z = 2(1+ 6); Normal: (@+ 3)y - z = 0

Miscellaneous Exercises (Page 117) 1 24z2 - y2 - 1502 + 2y + 224 = 0

2 17x2 - 2xy + 17y2 + 842 - 96y + 198 = 0

3 5 y 2 - 1 0 y - x + 4 = 0

(2-1)' ( Y + $ ) ~ 4 - 1 +

114 = 1; ellipse; centre (1, -3).

1 5 (y - 1)2 = -3(z - -); parabola; vertex (& , I ) .

3

6 ( Z + $ ~ - (y - f12 119 114

= 1; hyperbola; centre (-i, a).


7 Vertex (- 4 , O ) Equation of directrix X = -2 E x =.-i

Centre C(3, -2) Equations of directrices Y = f 5 t . . y = 3 o r y = - 7

Centre C(-$, 4) Asymptotes: Y = f &X - 1 = y = 5 f ++ + $1 Equations of directrices: X = f %

2 = x = ~ o r - 2

10 Tangent: y = 0; Normal: z = -1

11 Tangent: x $. y = 1; Normal: y - x + 1 = 0

12 a. 2 y - x + l = O ; x + p + 2 = O b. (2,;); (-4,2).

16 Tangent: 2g - 32 =, 12; Normal: 3y + 22 = 6.


Chapter 4

Exercises 4.1 (Page 124) 2 (i) a s i n O y + b c o s O x = a b and bcosOy-asinOz+a2e2sinOcosO=0

(ii) atanOy-bsecOx+ab=OandbsecOy+atan~x-a2e2secOtanO=0.

3 Equations of bisectors a t (x l , yl) are (A - B ) x + (A + B)y + C - k = 0 and (A + B)x - (A - B)y + C - k = 0 where A = ax1 + $by1 + g, B = abzl + cyl + f , C = gxl + fy l + d and k = (a - c ) z ~ y l + ;b(x: - y:) + gyl - f x l .

4 ( z + 2 a ) y 2 = - 4 a 3

6 The same as for the ellipse.


1 r = 2 1

e = - ellipse 1 + $c0sO1 2 '

2 3 - 3 r = e = 2, hvperbola 4 r = 2 hyperbola

1 - 2cosO' l + / c o s O 1 e = I '


1 , 6 r = Q 5 5 r = e = 1, parabola e = - hyperbola

l + - e 9 ~ + ~ c o s ~ ' 3 '

1 - 1 A 7 r = - where 6 = 6 - - e = h, hyperbola.

1 + f i s in(6- $) l + f i s i n d 4 '

Exercises 4.3 (Page 134) A

1 a = -. 7t2 - 3$ = 20 4 '

2 a = E . t2 + 5 $ - 4 t - 4 & = 0 6 ' A a = -. t2 - 4712 + 167 - 20 = 0 3 ' A

a = - - x 2 + f i v = 0 4 ' ?r a = -. 4 ' ~ ~ - ~ ~ - 2 f i ~ = 2

A = -' 4 ' 4 q 2 - 5 h t + f i q + 8 = ~

No. When b # 0 no rotation a can simultaneously make B = 0 and A = C. discriminant A < 0 ellipse

discriminant A > 0 hyperbola

discriminant A = 0 parabola.


Miscellaneous Exercises (Page '134)

4 r- = 1 2, r = 'I5 e = - ellipse e = 1, parabola

1 + i s i n e ' 5' 1 -s ine '

1 r = e = 4, hyperbola.

1 + JZsin(e + 5 ) '

9 discriminant A < 0 ellipse

10 discriminkt A > 0 hyperbola

Chapter 5

Exercises 5.2.2 (Page 150) 4 r(t) = (1, - , 2 ) + t(-12,4,3)

(i) t = 2 (ii) (3, -%, z )

(iii) h ( 5 + a, -15 - 3 6 , 1 0 + 2 6 ) and -&(-5 + ml 15 - 3K4,-10 + 2 a )


6 JAC1 = 106.2 km; C is in direction N41.72OW from A. If i is east and j is north then A>= -50&i+50(3- &) j

7 a. 57.69 m b, swims upstream inclined at 67.38' relative to the bank.

Exercises 5.2.4 (Page 159) 1 (i) - 2 i + 4 j - 1 0 k (ii) 1 6 i - 8 j - 1 6 k

2 ( p ~ q ) . r = O o r r = c u p + f I q w i t h c r = l , p = - 1

3 The set in (iii) is coplanar.

4 (i) 23 (ii) -102 (iii) -126

5 x + y = 3

6 x + z = 2

7 - x + y + 4 z = 4

8 ( r - a ) . ( b ~ c ) = O

9 (x, y, z) = A(-2, -1, l ) where X is any real number.

10 (x, y , z) = A(2,1,0) where A is any real number.

(i) cos-' (-$) = 112.39' (ii) cos-' (-$) = 91.82'

4 14 (i) % ( 4 i + 2 j - 3 k )

1 15 Component of c - a perpendicular to line is -(-I9 i + 11 j - 4 k). Foot of perpendicular

1 6

= -(I3 i + 7 j + 16 k). 6

Exercises 5.3 (Page 163) 1 (e' sin t + et cos t ) i + (e' cost - et sin t ) j + 3et k

2 s in t+ tcos t

3 2 s i n t i + ( 2 c o s t + 3 ) j + ( t s i n t - c o s t ) k 1 1

4 r(t) = -t2 i + -t3 k 2 3

5 Velocity v(t) = -2 sin t i + 2 cost j + k; speed = fi; acceleration a(t) = -2 cost i - 2 sin t j.

6 Displacement or position vector r(t) = (100t - 5t2) k; r = 500 k when velocity is 0.

7 a. d(-sin8,cosfI)

b. a(-sinOcos4 ~ - c o s 8 s i n $ ~ , - s i n 8 s i n $ ~ + c o s 8 c o s $ ~ , c o s 8 ~ )


1 r = - i + 3 k + ( 5 i + 3 j - 4 k ) t ; (+,%,0).

2 2 2 - 3 ~ + 4 2 = 1 3

4 (i) 8 (ii) 5 (iii) -23i+lOj-12k


6 a. scalar projection = - 3 ; vector projection = - i ( i - 2 j + 2k); angle between them

= cos-I - j) = 137.55O. ' ( 3 10

62 b. scalar projection = -T; vector projection = -%(-3 i + 4 k); angle between them = cos-I(-%) = 143.13O.

7 (i) x + 5 y + z = 5 , (ii) y + z = 3 .

8 (3,-1,-1)

9 -2sintcosti-(sin2t+2tcostsint)j+(sint+tcost+~os2t-sin2t)k

10 velocity = i + 4t j; acceleration = 4 j.

11 v = 6 c o s 2 t i - 6 s i n 2 t j + 2 k ; a = - 1 2 s i n 2 t i - l 2 c o s 2 t j

12 r ( t ) = (ut cosI9,utsinI9 - ig t2 ,0)

Chapter 6

Exercises 6.1 (Page 169) 1 velocity: 6t i + 2 j + 3t2 k; acceleration: 6 i + 6t j

2 velocity: u cos I9 i + (u sin 0 - gt) j; acceleration: -g j

3 velocity: pe-Pt i + (1 - pe'Pt) k ; acceleration: -p2e-Pt i + p2e-Pt k

5 b. s e c t t a n t i + s e c 2 t j .


1 Tangential component: 1 8 t / . \ / m ; normal component: -61 d m . 2 v = 2t; et = j; tangential component: 2; normal component: 0.

3 v = 1; et = - cost i - sint j; tangential component: 0; normal component: 1.

4 v = J T T G F ; e t = a i - 2 c t j 4c2t -2ac d-; tangential: a2 + 4c2t2

t/-; rmmal: d W = ' . 6 acceleration: a = (x, y) = (0, -288 cos 32).

Exercises 6.3 (Page 179) 7 (i) 720 m (ii) 12 + 5d8 sec (iii) 5 0 4 .

8 maximum height: 562.5 m; range: 2250 m.

Exercises 6.4 (Page 189) 500

1 6 ) (ii) 15 (iii) 75.

3k 4 (i) r -3 = 1 - -9

/I (ii) radial component: 2k3P - transverse component: 0.

- Answers to Exercises 257

59 5 (i) 12 m/sec (ii) - sec . (iii) 12 m/sec (iv) 8 m.

6 u

6 (i) r(t) = (- - sin a t , b cos 2at) a


1 b. v = - 2 c o s t i - s i n t j ; a = 2 s i n t i - c o s t j cost i- 2cos t j , 2 s i n t i + cost j

3 v = d c o s 2 t + 4 s i n 2 t ; e t = e n = tangential compe dcos2 t + 4 sin2 t dcos2 t + 4 sina t

3 sin t cos t nent : ; normal component: - 2

dcos2 t + 4 sin2 t dcos2 t + 4 sin2 t '

4 ~=i ;cosO-2 i6s inO-r~cosO-r62~osO;j i=~s inO+2i6s inO+r~cosO-r~2s inO

5 (0, - u 2 u 2 ~ s i n wx)

6 (i) 10 m/sec (ii) 110 - 1 0 4 sec (iii) 5fi m/sec.

7 velocity at (a, -b) = 0; velocity at (-a,O) = fabj.

k(m - - 8 (i) rm-" - "b+ 1 C1

(ii) radial component: nk2r2"-I - p2r2m-1; transverse component: kp(m + l)rm+""

1 0 0 d - 125 metres

Chapter 7


13 (i) -- 8 1

m/sec2 (ii) 1234.6 N

1 14 - N -

5

Exercises 7.2 (Page 198) 80 40

1 distance: - m velocity: - m/sec. 7 7

3 a. 3.72 sec b. 32.24 m/sec.

4 (i) g sin a (ii) gt sin a (iii) igt2sin a

(iv) time = d-, speed = d G


Exercises 7.3 (Page 212) 2 20.65" or 83.3g0


2 - = 0.253 m 4n2

1 3 0.4988% % -%

2 4 1.007 Hz

5 0.3444 sec; 0.365 m/sec.

6 2 d G da. Exercises 7.5 (Page 233)

25 - m/sec perpendicular to OA and making an angle of 60" with the velocity of B. 3

For A: rn2U vertically downwards and horizontally in the direction of its original motion. m l + rn2

For B: rn2U vertically downwards. m1+ m2

7 -- 25

(i) e = - rnl where ml , mz are masses of moving and stationary balls respectively. m2

(ii) r n l s r n z .

speed = dcos2 6 + e2 sin2 6 inclined at angle tan-'(e tan 6) with the horizontal.

(i) Formassrn: i(20fi-4fi)i+4fij;Formass2rn: 5 ( 5 4 + 8 f i ) i + 5 j

(ii) For mass rn: 5 f i i + 4 4 j ; For mass 2m: i ( 5 f i + 4 4 ) i + 5 j where i is taken along the line of centres.

Exercises 7.6 (Page 238) €

1 v* = -(cr - rnog) + ~6 ; cr - rnog > 0 is condition for rocket to rise r

m0c2 immediately. x = -(cr - mop) + 2r2


u2 1 u 1 1 -- - g ~ 2 above point of projection; time: - + -T.

29 8 9 2

2 maximum horizontal range: Ed-. 9

5 +U along line of centres; !U in direction making ahgle tan-' $ with line of centres.

6 e2"h.

Greek Alphabet 259

Greek Alphabet

Name

Alpha

Beta

Gamma

Delta

Epsilon

Zeta

Eta

Theta

Iota

Kappa

Lambda

Mu

-

Greek letter Name - Nu

Xi

Omicron

Pi

Rho

Sigma

Tau

Upsilon

Phi

Chi

Psi

Omega

260 Index

Index

acceleration vector 167 angle between two circles 64 angle between two lines 25

bisector of 33, 38 angle between two vectors 151 angular momentum 219 area of polygon 10 area of triangle 8 asymptotes 92 axes

rotation of 129 translation of 85

billiards 232 central orbit 186 centroid 7 chord of contact 59 circle

centre of 45 equation of 45 orthogonal 64 radius of 45 systems of 64 through three points 47

circumcircle of triangle 47 coaxial circles 70 coefficient of restitution 226 conic 77

parametric equations of 109 conic section 78 coordinate planes 135 cross product (see vector product) cycloid 110 cylindrical polar coordinates 171 determinant 8 direct impact 225 directed distance 136 direction cosine 140 directrix 78 displacement vector 165 distance between two points 1, 2

dot product (see scalar product) eccentricity 78 elastic limit 215 ellipse 77, 82, 88 equation of a straight line

intercept form 17, 19 parametric 22 slope-intercept form 17, 18 tangent form 17

equation of circle 45 standard form of 45

external division 2, 4, 5 focus 78 frame of reference 165 friction 194 gradient 12 homogeneous equation of second degree 37 Hooke's law 215 hyperbola 77, 82, 91 impulse 223 impulsive force 223 impulsive motion 223 internal division 2 intersection of circle and line 51 kinematics 165 line

distance of a point from 31 parallel 27 pencil of 35 perpendicular 27 vector equation of 145

major axis 88 median 6 midpoint 5 minor axis 88 moment of inertia 220 momentum 192 natural length 215 Newton's laws of motion 192 Newton's second law 141, 192

Index 261

normal circle 58 conic 100 curve 98, 172 plane 155

normal component null vector 137 oblique axes 129 oblique impact 225 orthogonal circles 64 parabola 77, 83, 87 parabolic motion 176 parallel axes 86 parallel lines 27 parallelogram law of vector addition 143 parametric equation of conic 109 particle 165 particle path 166 pencil of straight lines 35

vertex of 35 perfectly elastic object 226 perpendicular lines 27 plane polar coordinates 170 polar equation of a conic 124 polar of a point 60 pole 60 position vector 138 power of a point 68 projectile 199

maximum height of 200 range of 200

radial component 170 radical axis 68, 69 radius of curvature 173 rectilinear motion 176 reference point 165 reflecting property 118 reflecting telescope 123 rotation of axes 129 scalar 137 scalar product of two vectors 151

properties of 152 scalar triple product 158 second degree equation 128 simple pendulum 217 slope 12

speed 167 tangent

chord of contact 59 to circle 51, 52 to conic 100 to curve 98, 172 length of 57 pair of 55

tangential component 171 terminal speed 211 thrust 235 trajectory 166 translation of axes 85 transverse component 170 triangle law of vector addition 142 vector 137

addition 142, 145 parallelogram law of 143 properties of 143 triangle law of 142

component of 139, 147 equality of 139 final point of 137 initial point of 137 line segment representation of 137 magnitude of 137, 140 orthogonal 152 parallel 141 position 138 scalar component of 153 unit 141 vector component of 153

vector function 160 continuity of 160 derivative of 161 integration of 162 limit of 160

vector product of two vectors 156 properties of 156

vector triple product 158 velocity vector 167 vertex of ellipse 88 vertex of hyperbola 94 Young's modulus of elasticity 216 zero vector 137

' 4 .

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