preliminary mathematics extension 1 - dux collegemany problems require you to find the intersection...
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Preliminary
Mathematics
Extension 1
Parametric Equations
Term 1 โ Week 1
Name โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ.
Class day and time โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ
Teacher name โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ...
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Preliminary Mathematics Extension 1 - Parametric Equations
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Term 1 โ Week 1 โ Theory
CARTESIAN REPRESENTATION OF THE PARABOLA ๐๐ = ยฑ๐๐๐
(REVISION):
A parabola is defined as the locus of all points that are equidistant from a fixed point and a given line. The
point is known as the โfocusโ and the line is known as the โdirectrixโ.
The parabola with focus (0, ๐) and directrix ๐ฆ = โ๐ has Cartesian equation ๐ฅ2 = 4๐๐ฆ.
The parabola with focus (0, โ๐) and directrix ๐ฆ = ๐ has Cartesian equation ๐ฅ2 = โ4๐๐ฆ.
EXAMPLE:
Find the locus of a point P that moves so that its distance from the point (0, 2) is the same as its distance from
the line ๐ฆ = โ2.
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SOLUTION:
Let the point be ๐(๐ฅ, ๐ฆ)
Distance from ๐ to (0, 2) is given by โ(๐ฅ โ 0)2 + (๐ฆ โ 2)2 = โ๐ฅ2 + (๐ฆ โ 2)2
Distance from ๐ to ๐ฆ = โ2 is given by |0๐ฅ+1๐ฆ+2
โ02+12| = |
๐ฆ+2
1| = |๐ฆ + 2|
โด โ๐ฅ2 + (๐ฆ โ 2)2 = |๐ฆ + 2|
๐ฅ2 + (๐ฆ โ 2)2 = (๐ฆ + 2)2
๐ฅ2 + ๐ฆ2 โ 4๐ฆ + 4 = ๐ฆ2 + 4๐ฆ + 4
๐ฅ2 = 8๐ฆ
The locus is ๐ฅ2 = 8๐ฆ.
PARAMETRIC EQUATION OF THE PARABOLA ๐๐ = ยฑ๐๐๐
The parabola ๐ฅ2 = 4๐๐ฆ can be represented by the parametric equations:
๐ฅ = 2๐๐ก and ๐ฆ = ๐๐ก2
where ๐ก is known as a โparameterโ.
Therefore ๐(2๐๐, ๐๐2) would represent a general point on the parabola, and ๐(2๐๐, ๐๐2) would represent
another point on the parabola.
This means that substituting in any given value for the parameter will give us exactly one point on the
parabola, and the locus of all such points is the parabola.
For example, consider the parabola ๐ฆ = ๐ฅ2, or ๐ฅ2 = 4 (1
4) ๐ฆ (i.e. ๐ =
1
4 )
If we let ๐ก = 2, we obtain the point (2 (1
4) (2), (
1
4) (2)2) = (1, 1)
If we let ๐ก = โ4, we obtain the point (2 (1
4) (โ4), (
1
4) (โ4)2) = (โ2, 4)
The use of parametric representations allows important properties of the parabola and the equations of
related curves (e.g. tangents, normals) to be expressed as functions of one parameter, ๐ก. This is helpful
because it simplifies the algebra involved.
In the following sections, both the parametric and Cartesian representations are used to derive the equations
of tangents, normals and chords.
Most questions will ask you to derive one or more of these equations in the first step, so that the results can
be used to prove further properties. Therefore it is useful to know both the derivations and the results. These
can be learnt simply through practicing on sample questions. There is no need to rote learn them.
TANGENTS TO THE PARABOLA ๐๐ = ๐๐๐
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1. Cartesian Representation
Let ๐(๐ฅ1, ๐ฆ1) be a point on the parabola ๐ฅ2 = 4๐๐ฆ.
๐ฅ2 = 4๐๐ฆ
Differentiating both sides with respect to ๐ฅ,
2๐ฅ = 4๐.๐๐ฆ
๐๐ฅ
โด๐๐ฆ
๐๐ฅ=
๐ฅ
2๐
โด at the point ๐(๐ฅ1, ๐ฆ1),
๐๐ฆ
๐๐ฅ=
๐ฅ1
2๐
So the tangent is given by
๐ฆ โ ๐ฆ1 =๐ฅ1
2๐(๐ฅ โ ๐ฅ1)
โด 2๐๐ฆ โ 2๐๐ฆ1 = ๐ฅ๐ฅ1 โ ๐ฅ12
โด ๐ฅ๐ฅ1 = 2๐๐ฆ โ 2๐๐ฆ1 + ๐ฅ12
= 2๐๐ฆ โ 2๐๐ฆ1 + 4๐๐ฆ1 as (๐ฅ1, ๐ฆ1) lies on ๐ฅ2 = 4๐๐ฆ.
โด ๐ฅ๐ฅ1 = 2๐(๐ฆ + ๐ฆ1)
โด The tangent to the parabola ๐ฅ2 = 4๐๐ฆ at a point ๐(๐ฅ1, ๐ฆ1) is given by
๐ฅ๐ฅ1 = 2๐(๐ฆ + ๐ฆ1)
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2. Parametric Representation
Let ๐(2๐๐, ๐๐2) be a point on the parabola ๐ฅ2 = 4๐๐ฆ with parameter ๐.
๐ฅ2 = 4๐๐ฆ
Differentiating both sides with respect to ๐ฅ,
2๐ฅ = 4๐.๐๐ฆ
๐๐ฅ
โด ๐๐ฆ
๐๐ฅ=
๐ฅ
2๐
โด at the point ๐(2๐๐, ๐๐2),
๐๐ฆ
๐๐ฅ=
2๐๐
2๐= ๐
So the tangent is given by
๐ฆ โ ๐๐2 = ๐(๐ฅ โ 2๐๐)
โด ๐ฆ โ ๐๐2 = ๐๐ฅ โ 2๐๐2
โด ๐ฆ = ๐๐ฅ โ ๐๐2
โด The tangent to the parabola ๐ฅ2 = 4๐๐ฆ at a point ๐(2๐๐, ๐๐2) is given by
๐ฆ = ๐๐ฅ โ ๐๐2
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NORMALS TO THE PARABOLA ๐๐ = ๐๐๐
1. Cartesian Representation
Let ๐(๐ฅ1, ๐ฆ1) be a point on the parabola ๐ฅ2 = 4๐๐ฆ.
๐ฅ2 = 4๐๐ฆ
Differentiating both sides with respect to ๐ฅ,
2๐ฅ = 4๐.๐๐ฆ
๐๐ฅ
โด๐๐ฆ
๐๐ฅ=
๐ฅ
2๐
โด at the point ๐(๐ฅ1, ๐ฆ1), the gradient of the tangent to the curve is given by
๐๐ฆ
๐๐ฅ=
๐ฅ1
2๐
โด the gradient of the normal is โ2๐
๐ฅ1
โด the normal to the parabola ๐ฅ2 = 4๐๐ฆ at a point ๐(๐ฅ1, ๐ฆ1) is given by
๐ฆ โ ๐ฆ1 = โ2๐
๐ฅ1
(๐ฅ โ ๐ฅ1)
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2. Parametric Representation
Let ๐(2๐๐, ๐๐2) be a point with parameter ๐ on the parabola ๐ฅ2 = 4๐๐ฆ.
๐ฅ2 = 4๐๐ฆ
Differentiating both sides with respect to ๐ฅ,
2๐ฅ = 4๐.๐๐ฆ
๐๐ฅ
โด ๐๐ฆ
๐๐ฅ=
๐ฅ
2๐
โด at the point ๐(2๐๐, ๐๐2), the gradient of the tangent to the curve is given by
๐๐ฆ
๐๐ฅ=
2๐๐
2๐= ๐
โด the gradient of the normal is โ1
๐
So the normal is given by
๐ฆ โ ๐๐2 = โ1
๐(๐ฅ โ 2๐๐)
โด ๐๐ฆ โ ๐๐3 = โ๐ฅ + 2๐๐
โด ๐ฅ + ๐๐ฆ = ๐๐3 + 2๐๐
โด the normal to the parabola ๐ฅ2 = 4๐๐ฆ at a point ๐(2๐๐, ๐๐2) is given by
๐ฅ + ๐๐ฆ = ๐๐3 + 2๐๐
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INTERSECTION OF TANGE NTS AND NORMALS OF THE PARABOLA ๐๐ = ๐๐๐
Many problems require you to find the intersection of the tangents or normals at point ๐(2๐๐, ๐๐2) and ๐(2๐๐, ๐๐2), and then to prove some property involving this intersection. Thus is it worthwhile to know both what the point is, and how to derive it.
1. Intersection of Tangents
Let the intersection be ๐
Equation of the tangent at ๐: ๐ฆ = ๐๐ฅ โ ๐๐2 โ โ โ โ(1)
Equation of the tangent at ๐: ๐ฆ = ๐๐ฅ โ ๐๐2 โ โ โ โ(2)
Solving (1) and (2) simultaneously,
๐๐ฅ โ ๐๐2 = ๐๐ฅ โ ๐๐2
๐๐ฅ โ ๐๐ฅ = ๐๐2 โ ๐๐2
(๐ โ ๐)๐ฅ = ๐(๐ + ๐)(๐ โ ๐)
โด ๐ฅ = ๐(๐ + ๐) as ๐ โ ๐ and so ๐ โ ๐ โ 0
โด ๐ฆ = ๐๐ฅ โ ๐๐2
= ๐๐(๐ + ๐) โ ๐๐2
= ๐๐(๐ + ๐ โ ๐)
= ๐๐๐
โด ๐[๐(๐ + ๐), ๐๐๐] is the intersection of the tangents at points ๐ and ๐.
2. Intersection of Normals
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Let the intersection be ๐
Equation of normal at ๐: ๐ฅ + ๐๐ฆ = ๐๐3 + 2๐๐ โ โ โ โ(1)
Equation of normal at ๐: ๐ฅ + ๐๐ฆ = ๐๐3 + 2๐๐ โ โ โ โ(2)
Subtracting (2) from (1):
๐๐ฆ โ ๐๐ฆ = ๐๐3 + 2๐๐ โ ๐๐3 โ 2๐๐
(๐ โ ๐)๐ฆ = ๐(๐3 โ ๐3 + 2(๐ โ ๐))
= ๐(๐ โ ๐)(๐2 + ๐๐ + ๐2 + 2)
โด ๐ฆ = ๐(๐2 + ๐๐ + ๐2 + 2) as ๐ โ ๐ and so ๐ โ ๐ โ 0
โด ๐ฅ = ๐๐3 + 2๐๐ โ ๐๐ฆ
= ๐๐3 + 2๐๐ โ ๐๐(๐2 + ๐๐ + ๐2 + 2)
= ๐๐3 + 2๐๐ โ ๐๐3 โ ๐๐2๐ โ ๐๐๐2 โ 2๐๐
= โ๐๐2๐ โ ๐๐๐2
= โ๐๐๐(๐ + ๐)
โด ๐[โ๐๐๐(๐ + ๐), ๐(๐2 + ๐๐ + ๐2 + 2)] is the intersection of the normals at points ๐ and ๐.
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Term 1 โ Week 1 โ Homework 1. By using differentiation, find the equation of the tangent to the parabola at the indicated points:
a) tx 2= ,2ty = at the point where 1=t
b) tx 4= ,22ty = at the point where
2
1โ=t
c) tx = ,2
2
1ty = at the point where 4=t
d) atx 2= ,2aty = at the point where 3=t
e) yx 42 = at the point ( )1,2โ
f) yx 82 โ= at the point
โโ
2
1,2
g) yx 62 = at the point ( )6,6
h) ayx 42 = at the point ( )11, yx
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2.
(i) Find the equation of the tangent to the parabola yx 82 = at the point ( )22,4 tt
(ii) Hence determine all tangents to the parabola that pass through the point ( )1,1 โ
3. ( )2,2 ppP and ( ) ( )( )211 ,2pp
Q are two variable points on the parabola yx 42 = . The tangents at P and
Q intersect at a point T.
(i) Find the equation of the tangent to the parabola at P.
(ii) Determine the coordinates of T.
(iii) Hence find the Cartesian equation of the locus of T.
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4. The line 1=+byax is tangent to the parabola yx 42 โ= . Find the conditions on ๐ and ๐.
5. ( )2,2 apapP is a variable point on the parabola ayx 42 = . The tangent at P intersects the x-axis at A
and the y-axis at B. C is the fourth vertex of rectangle OACB.
(i) Find the coordinates of C in terms of ๐.
(ii) Hence show that the locus of C is a parabola and state its vertex and focus.
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6. ( )2,2 apapP is a variable point on the parabola ayx 42 = . T is the foot of the perpendicular drawn
from the focus to the tangent at P. Find the Cartesian equation of the locus of T.
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7. By using differentiation, find the equation of the normal to the parabola at the indicated points:
a) tx 2= ,2ty = at the point where 2โ=t
b) tx 6= ,23ty = at the point where 4=t
c) tx = ,2
2
1ty = at the point where 1=t
d) atx 2= ,2aty = at the point where pt =
e) yx 42 = at the point ( )1,2
f) yx โ=2 at the point ( )1,1 โ
g) yx 122 = at the point ( )3,6โ
h) yx4
12 = at the point ( )11, yx
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8.
(i) Find the equation of the normal to the parabola atx 2= ,2aty = at the point where pt = .
(ii) The normal intersects the ๐ฅ-axis at A and the ๐ฆ-axis at B. Find the coordinates of A and B.
(iii) Hence determine the area of AOB
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9.
(i) Find the equation of the parabola that is symmetrical about the y-axis and passes through the points
( )1,1 and ( )4,4โ .
(ii) Find the normal to the parabola at the point ( )1,1 .
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10. ( )2,2 apapP is a variable point on the parabola ayx 42 = . The normal at P intersects the y-axis at T. M
is the midpoint of PT.
(i) Find the coordinates of T.
(ii) Hence find the coordinates of M and determine the Cartesian equation of the locus of M.
End of Homework