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Polar Conic to Cartesian

  1. Aug 16, 2009 #1
    1. The problem statement, all variables and given/known data
    The equation of a conic in polar coordinates is:
    [tex] r = \frac{r_o}{1-\epsilon cos(\theta)}[/tex].

    [tex]\epsilon [/tex] is the eccentricity, 0 for a circle, (0,1) for an ellipse, 1 for a parabola, and >1 for a hyperbola.

    What is this equation expressed in Cartesian coordinates?

    2. Relevant equations
    [tex]x=r cos(\theta)[/tex]

    3. The attempt at a solution
    [tex] r = \frac{r_o}{1-\epsilon cos(\theta)}[/tex]

    [tex] r (1-\epsilon cos(\theta)) = r_o[/tex]

    [tex] r - r cos(\theta) \epsilon = r_o[/tex]

    [tex] r - x \epsilon = r_o[/tex]

    [tex] r = r_o + x \epsilon[/tex]

    [tex] (r)^2 = (r_o + x \epsilon)^2 [/tex]

    [tex] x^2 + y^2 = r_o^2 + 2 (x \epsilon) r_o + (x \epsilon)^2 [/tex]

    [tex] x^2 - (x \epsilon)^2 - 2 \epsilon r_o x + y^2 = r_o^2 [/tex]

    [tex] (1 - \epsilon^2)x^2 - 2 \epsilon r_o x + y^2 = r_o^2 [/tex]

    My textbook says this equation should be:

    [tex] (1 - \epsilon^2)x^2 - 2 r_o x + y^2 = r_o^2 [/tex]

    (Notice that we differ on the coefficient of x. Is the textbook missing an epsilon by a misprint? or did I mess up somewhere?

    Thanks for the help.
  2. jcsd
  3. Aug 16, 2009 #2


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    Homework Helper

    Hi Knissp! :smile:

    (have an epsilon: ε :wink:)

    No, you're correct …

    if we put ε = 0, for a circle, then your equation is obviously correct, but the book's equation is (x - r0)2 + y2 = 2r02, which is a circle, but not the right one! :wink:
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