Polar Equations of Conics
description
Transcript of Polar Equations of Conics
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Polar Equations of Conics
Combining Skills We Know (10.6)
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POD
Give a 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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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
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yx
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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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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Try it
Rewrite the equation of this conic. Finish the algebra.
Are we good?
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8010091685
1009405
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1023
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yx
yx
yxx
yxx
xxyx
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xyx
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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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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
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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
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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
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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
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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
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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
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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
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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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93636
36
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yx
yx
yxx
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xxyx
xyx