Unit 8.5

8.5
Polar Equations
of Conics
Copyright © 2011 Pearson, Inc.

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.
Copyright © 2011 Pearson, Inc. Slide 8.5 - 2

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
Given that the focus is at the pole, write a polar equation
for the conic with eccentricity 4/5 and directrix x  3.
Setting e  4 / 5 and k  3 in r 
ke
1 ecos
yields
r 
34 / 5
1 4 / 5cos

12
5  cos

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
Determine the eccentricity, the type of conic,
and the directrix. r 
6
3 2cos
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.

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.

r 
6
3 2cos
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.

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
  2 1
a 
e
e 
1 cos
r



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)

Unit 8.5

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Unit 8.5