Unit 8.5

Copyright © 2011 Pearson, Inc.

8.5

Polar Equations

of Conics

Copyright © 2011 Pearson, Inc. Slide 8.5 - 2

What you’ll learn about

Eccentricity Revisited

Writing Polar Equations for Conics

Analyzing Polar Equations of Conics

Orbits Revisited

… and why

You will learn the approach to conics used by

astronomers.


Focus-Directrix Definition Conic

Section

A conic section is the set of all points in a

plane whose distances from a particular point

(the focus) and a particular line (the directrix)

in the plane have a constant ratio. (We assume

that the focus does not lie on the directrix.)


Focus-Directrix Eccentricity

Relationship

If P is a point of a conic section, F is the conic's focus,

and D is the point of the directrix closest to P, then

e PF

PD and PF e PD, where e is a constant and

the eccentricity of the conic. Moreover, the conic is

g a hyperbola if e 1,

g a parabola if e 1,

g an ellipse if e 1.


The Geometric Structure of a Conic

Section


A Conic Section in the Polar Plane


Three Types of Conics for r =

ke/(1+ecosθ)


Polar Equations for Conics


Example Writing Polar Equations of

Conics

Given that the focus is at the pole, write a polar equation

for the conic with eccentricity 4/5 and directrix x 3.


Example Writing Polar Equations of

Conics

Setting e 4 / 5 and k 3 in r ke

1 ecos yields

r 3 4 / 5

1 4 / 5 cos

12

5 cos

Given that the focus is at the pole, write a polar equation

for the conic with eccentricity 4/5 and directrix x 3.


Example Identifying Conics from

Their Polar Equations

Determine the eccentricity, the type of conic,

and the directrix. r 6

3 2cos


Example Identifying Conics from

Their Polar Equations

Divide the numerator and the denominator by 3.

r 2

1 (2 / 3)cos

The eccentricity is 2/3 which means the conic is an ellipse.

The numerator ke 2 (2 / 3)k, so k 3

and the directrix is y 3.

Determine the eccentricity, the type of conic,

and the directrix. r 6

3 2cos


Example Analyzing a Conic

Analyze the conic section given by the equation

r 6

3 2cos

Include in the analysis the values of e, a, b, and c.



Divide the numerator and the

denominator by 3 yields

r 6

1 2cos

The eccentricity is e 2, and

thus the conic is a

hyperbola, shown here.

r

6

3 2cos



The vertices have polar coordinates (2, 0) and (Ğ6, ).

So 2a 6 2 4, and thus a 2

The vertex (2, 0) is 2 units to the

right of the pole, the pole is the

focus of the hyperbola,

so c a 2, and thus c 4.

r

6

3 2cos



To find b, use Pythagorean relation:

b c2 a2 16 4 12 2 3

With all the information, we can

write the Cartesian equation of

the hyperbola:

x 4 2

4

y2

12 1


Semimajor Axes and Eccentricities of

the Planets


Ellipse with Eccentricity e and

Semimajor Axis a

21

1 cos

a er

e


Quick Review

1. Solve for r. (4,) (r, )

2. Solve for . (3, 5 /3)=( 3,), 2 2

3. Find the focus and the directrix of the parabola.

x2 12y

Find the focus and the vertices of the conic.

4. x2

16

y2

9 1

5. x2

9

y2

16 1


Quick Review Solutions

1. Solve for r. (4,) (r, ) 4

2. Solve for . (3, 5 /3)=( 3,), 2 2 4 / 3

3. Find the focus and the directrix of the parabola.

x2 12y (0,3); y 3

Find the focus and the vertices of the conic.

4. x2

16

y2

9 1 (5,0); ( 4,0)

5. x2

9

y2

16 1 (0, 7); (0, 4)

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Unit 8.5

Education