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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.

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Polar Equations of Conics

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  1. Polar Equations of Conics Combining Skills We Know (10.6)

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

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

  4. 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?

  5. 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?

  6. 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.

  7. 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.

  8. 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?)

  9. 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?

  10. 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.

  11. 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?

  12. 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).

  13. 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?

  14. 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.

  15. 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?

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

  17. Try it Rewrite the equation of this conic. Finish the algebra. Are we good?

  18. Try it Rewrite the equation of this conic. Does it match? a = 6, b = √20, center (-4, 0). Yeah, we’re good.

  19. 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?

  20. 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

  21. 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.)

  22. 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.)

  23. 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.

  24. 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

  25. 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

  26. 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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