polar equations of conics

Polar Equations of Conics

Combining Skills We Know (10.6)

POD

Give a rectangular equation for an ellipse centered on (-4,0), having an x-radius of 6 and a y radius of √20.

POD

Give the rectangular equation for an ellipse centered on (-4,0), having an x-radius of 6 and a y radius of √20.

1

2036

4 22

yx

POD

Give the rectangular equation for an ellipse centered on (-4,0), having an x-radius of 6 and a y radius of √20.

What are the foci? Any idea how to write this in polar equation form?

1

2036

4 22

yx

The General Form

The forms for conic graphs in polar equation form, with a focus at the pole:

d = distance from focus to directrix

e = eccentricity = c/a parabola: e = 1

ellipse: e < 1

hyperbola: e > 1

Which of these do you expect to orient vertically and which horizontally?

cos1 e

der

sin1 e

der

Try it– use polar graph paper

Sketch the graph of this conic. First step: rewrite into the given form, which means the denominator leads off with 1.

cos23

10

r

Try it

Sketch the graph of this conic. First step: rewrite into the given form.

In this form, we can tell that e = 2/3, which is less than 1, so it’s an ellipse. With a focus at the pole, and oriented horizontally.

cos32

1

310

cos23

10

r

Try it

Sketch the graph of this conic. Second step: set θ = 0 and θ = π to get the vertices. (Why?)

(If it were sinθ, and oriented vertically, we’d use

θ = π/2 and θ = 3π/2 to find the vertices. Why?)

cos23

10

r

Try it

Sketch the graph of this conic. Second step: set θ = 0 and θ = π to get the vertices.

If θ = 0, r = 2. If θ = π, r = 10. So the points (2, 0) and (10, π) are the vertices.

What is the center? Think.

What is the focal length?

cos23

10

r

Try it

Sketch the graph of this conic. Second step: set θ = 0 and θ = π to get the vertices.

If θ = 0, r = 2. If θ = π, r = 10. So the points (2, 0) and (10, π) are the vertices.

What is the center? (4, π) So, the focal length is 4.

cos23

10

r

Try it

Sketch the graph of this conic. Third step: graph the points.

The points (2, 0) and (10, π) are the vertices.The center is (4, π).One focus is (0, 0). What’s the other?

cos23

10

r

Try it

Sketch the graph of this conic. Third step: graph the points.

The points (2, 0) and (10, π) are the vertices.The center is (4, π).One focus is (0, 0). What’s the other? (8, π).

cos23

10

r

Try it

Before we rewrite the equation of this conic, let’s anticipate a few elements.

What is the long axis? How is it oriented?

So, what radius do we know?

cos23

10

r

Try it


What is the long axis? 12 How is it oriented?

So, what radius do we know? x-radius = 6

We use a = 6, and e = c/a to find b.

cos23

10

r

Try it


We use a = 6, and e = c/a to find b.

e = 2/3 = c/6 c = (2/3)6 = 4 (Didn’t we already determine this?)

a2 = b2 + c2 62 = b2 + 42 b = √20

Do we use sound mathematical reasoning?

cos23

10

r

Try it

Rewrite the equation of this conic. Use the elements we’ve used already.

1023

10cos23

10cos23

cos23

10

22

xyx

rr

r

r

Try it

Rewrite the equation of this conic. Finish the algebra.

Are we good?

1

2036

4

180945

8010091685

1009405

44010099

2109

2103

1023

22

22

22

22

222

222

22

22

yx

yx

yxx

yxx

xxyx

xyx

xyx

xyx

Try it

Rewrite the equation of this conic. Does it match?

a = 6, b = √20, center (-4, 0). Yeah, we’re good.

1

2036

4 22

yx

Want another?

Try a different conic.

Sketch it first.

Find the eccentricity.

What shape is it?

How is it oriented?

So, what angles to use to find the vertices?

cos62

12

r

Want another?

Try a different conic.

Find the eccentricity. 3

What shape is it? hyperbola

How is it oriented? horizontally

So, what angles to use to find the vertices? 0 and π

cos31

6

cos62

12

r

r

Want another?

Horizontal hyperbola

Vertices: θ = 0 r = -3 (-3, 0)

θ = π r = 3/2 (3/2, π)

Focus at (0,0)

Center where? (There are a couple of ways to find it.)

cos31

6

r

Want another?


Vertices: (-3, 0), (3/2, π)

Focus: (0,0)

Center (using average of vertices): (-9/4,0)

(We could also have found c, the focal length, and subtracted it from the pole.)

cos31

6

r

49

22

3

22

33

Want another?


Vertices: (-3, 0), (3/2, π)

Focus: (0,0)

Center: (-9/4,0)

We still need a and b in order to draw the box for the hyperbola.

cos31

6

r

Want another?


2a = the length between vertices = 3/2 a = ¾

e = 3 = c/a 3 = c/(¾) c = 3(¾) = 9/4

b2 = c2 – a2 = 81/16 – 9/16 b = 3/√2 ≈ 2.12

= 72/16 = 9/2

cos31

6

r

Want another?


Rewrite to check--

set up:

Vertices: (-3, 0), (3/2, π)

Focus: (0,0)

Center: (-9/4,0)

a = ¾ b = 3/√2

xyx

rr

rr

r

r

36

cos36

6cos3

6cos31

cos31

6

22

Want another?


Rewrite to check--

finish with lots

of fractions:

Vertices: (-3, 0), (3/2, π)

Focus: (0,0)

Center: (-9/4,0)

a = ¾ b = 3/√2

We’re good.

1

29

169

49

29

498

2813616

812

98

36368

93636

36

22

22

22

22

222

22

yx

yx

yxx

yxx

xxyx

xyx

polar equations of conics

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