chapter 6: analytic geometry


Chapter 6: Analytic Geometry

6.1 Circles and Parabolas

6.2 Ellipses and Hyperbolas

6.3 Summary of the Conic Sections

6.4 Parametric Equations


• Conic sections presented in this chapter are of the form

where either A or C must be nonzero.


,022 FExCyDxAx

Conic Section Characteristic Example

Parabola

Circle

Ellipse

Hyperbola

Either A = 0 or C = 0, but not both.

y = x2

x = 3y2 + 2y – 4

0CA 1622 yx0, ACCA

0AC

125

2

16

2 yx

122 yx



Last 3 rows of the Table on pg 6-52


6.3 Determining Type of Conic Section

• To recognize the type of conic section, we may need to transform the equation.

Example Decide on the type of conic section represented by each equation.

(a)

(b)

(c)

(d)

22 525 yx 61549164 22 yyxx

41108 22 yyxx

07862 yxx



Solution (a)

This equation represents a hyperbola centered at the origin.

1525

255525

22

22

22

yxyx

yx



(b) Coefficients of the x2- and y2-terms are unequal and both positive. This equation may be an ellipse. Complete the square on x and y.

This is an ellipse centered at (2, –3).

14

)3(9

)2(

36)3(9)2(4

6181)96(916)44(4

61)996(9)444(4

61)6(9)4(4

22

22

22

22

22

yx

yx

yyxx

yyxx

yyxx



(c) Coefficients of the x2- and y2-terms are both 1. This equation may be a circle. Complete the square on x and y.

This equation is a circle with radius 0; that is, the point (4, –5).

0)5()4(

251641)2510()168(

41108

22

22

22

yx

yyxx

yyxx



(d) Since only one variable is squared, x2, the equation represents a parabola. Solve for y (the variable that is not squared) and complete the square on x (the squared variable).

The parabola has vertex (3, 2) and opens downward.

2)3(81

16)3(8

97)96(8

768

0786

2

2

2

2

2

xy

xy

xxy

xxy

yxx


6.3 Eccentricity

• The constant ratio is called the eccentricity of the conic, written e.

A conic is the set of all points P(x, y) in a plane such that the ratio of the distance from P to a fixed point and the distance from P to a fixed line is constant.


6.3 Eccentricity of Ellipses and Hyperbolas

• Ellipses and hyperbolas have eccentricity

where c is the distance from the center to a focus.

• For ellipses, a2 > b2 and

• Note that ellipses with eccentricity close to 0 have a circular shape because b a as c 0.

,ac

e

So, .22 bac

)hyperbolasfor 1(.10

10

0

eeac

ac


6.3 Finding Eccentricity of an Ellipse

Example Find the eccentricity of

Solution Since 16 > 9, let a2 = 16, giving a = 4.

.1169

22

yx

7916

22

bac

66.47

ac

e


6.3 Figures Comparing Different Eccentricities of Ellipses and Hyperbolas


6.3 Finding Equations of Conics using Eccentricity

Example Find an equation for each conic with center at the origin.

(a) Focus at (3, 0) and eccentricity 2(b) Vertex at (0, –8) and e =

Solution(a) e = 2 > 1, this conic is a hyperbola with c = 3.

21



The focus is on the x-axis, so the x2-term is positive. The equation is

427

49

923

3

. find to Use

23

32

222

2222

b

bacb

a

a

ac

e

.1274

94

or1

427

49

2222

yxyx



(b) Since the conic is an ellipse. The vertex at (0, –8) indicates that the vertices lie on the y-axis and a = 8.

Since the equation is

,121 e

c

cac

e

482

1

,481664222 cab

.14864

22

xy


6.3 Applying an Ellipse to the Orbit of a Planet

Example The orbit of the planet Mars is an ellipse with the sun at one focus. The eccentricity of the ellipse is 0.0935, and the closest Mars comes to the sun is 128.5 million miles. Find the maximum distance of Mars from the sun.

SolutionUsing the given figure, Mars is

- closest to the sun on the right - farthest from the sun on the left

Therefore, - smallest distance is a – c - greatest distance is a + c


6.3 Applying an Ellipse to the Orbit of a Planet

Since a – c = 128.5, c = a – 128.5. Using e = 0.0935, we find a.

The maximum distance of Mars from the sun is about 155.1 million miles.

128.50.0935

0.0935 128.5141.8

ce

aa

aa aa

1.1553.138.141 3.135.1288.141

cac

chapter 6: analytic geometry

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