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Polar equations of conics question

  1. Jun 11, 2009 #1
    for this problem, i've been given the vertices of the hyperbola as (4, pi/2) and (-1, 3pi/2). the question asks to find the polar equation of this hyperbola.

    so what i did was do a quick sketch of the graph. (4, pi/2) is essentially (0,4) and (-1, 3pi/2) is essentially (0,1). the midpoint of the two vertices is (0,2.5) so that's the center. since the focus is at (0,0), c = 2.5 and a = 1.5

    therefore e = c/a and e = 5/3

    so i wrote out the equation r = (5/3)p / 1 + (5/3)sin& since the directrix is above the x axis or polar axis between the two vertices. (by the way i'm using "&" to stand for the greek letter theta.)

    then i just used one of the points they gave me (4, pi/2) and plugged it into the equation to solve for p. 4 = (5/3)p / (1 + (5/3)) when i solve for p, i get that p = 32/5.

    however when i plug in the other point they gave me for the hyperbola (-1, 3pi/2), i get a different p value. -1 = (5/3)p / (1 - (5/3)) and when i solve for p i get that p = 2/5.

    what's going on? how come the p values are different? both points are vertices of the hyperbola so i don't know why i'm getting different answers. the back of my textbook says that p = 8/5. how did the book get that? and how am i getting two different p values? please help me! i'm so confused.
     
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  3. Jun 12, 2009 #2

    HallsofIvy

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    What are you using for the eccentricity? Assuming the vertices are also given in polar coordinates, the vertices are at (0, 4) and (0, -1) so the center is at (0,3/2) but there are an infinite number of hyperboles that fit that.
     
  4. Jun 12, 2009 #3
    the vertices they have given me are (4, pi/2) and (-1, 3pi/2) which in rectangular coordinates is (0,4) and (0,1) so the center of the hyperbola is (0,5/2) and (0,0) is one focus of the hyperbola.

    so using that info i determined that c = 2.5 (the distance from center to focus) and a = 1.5 (distance from center to vertex). and i know that e = c/a, so i did e = 2.5 / 1.5 = 5/3

    so my eccentricity of this hyperbola is 5/3.
     
  5. Jun 12, 2009 #4

    gabbagabbahey

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    Hmmm... shouldn't this be:

    [tex]r=\frac{\left(\frac{5}{3}\right)p}{-1+\left(\frac{5}{3}\right)\sin(\theta)}[/tex]


    ???:wink:
     
  6. Jun 18, 2009 #5
    hm in my textbook, they only show the formulas r = ep / (1 +/- cos(&)) and r = ep / (1 +/- sin(&))

    (& stands for "theta")

    how did you get the -1 in your equation?
     
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