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How do I express the Kepler general orbit $r(\phi)$ in rectangular coordinates?

I use the identities $x=r\cos\phi$, $y=r\sin\phi$, and $r^2 = x^2 + y^2$, but I block at some point.

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Remember that the Kepler orbit is

$$ r = {1\over A + B cos(\theta)} $$


$$ Ar + B r cos(\theta) =1 $$


$$ A \sqrt{x^2 + y^2} + B x = 1 $$

$$ A \sqrt{x^2 + y^2} = 1- Bx $$

$$ A^2 (x^2 + y^2) = 1 - 2 Bx + B^2 x^2 $$

You can factor the x equation by completing the square, if you want to find the center of the ellipse.

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Of course you've assumed here that the perihelion of the orbit is on the x axis. But then, the general case can be obtained just as easily by using the addition formula for $\cos(\phi-\phi_0)$. – celtschk Jan 16 '12 at 7:01
@celtschk: does this need an edit? You can figure it out by a moment's reflection, and a general axis makes the formulas annoying unless you rotate the axis to x anyway. – Ron Maimon Jan 16 '12 at 7:54
I don't think you have to do the general calculation; however I think it is important that the reader is aware that the calculation is for a special case, hence my comment. – celtschk Jan 16 '12 at 8:37

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