polar equations of conics
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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.
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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?
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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
Before we rewrite the equation of this conic, let’s anticipate a few elements.
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
Before we rewrite the equation of this conic, let’s anticipate a few elements.
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?
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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.
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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?
Horizontal hyperbola
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?
Horizontal hyperbola
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?
Horizontal hyperbola
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?
Horizontal hyperbola
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?
Horizontal hyperbola
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.
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169
49
29
498
2813616
812
98
36368
93636
36
22
22
22
22
222
22
yx
yx
yxx
yxx
xxyx
xyx
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