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So I'm looking to derive the equation of a hyperbolic orbit from the general expression for a conic section $$r=\frac{l}{e\cos\theta+1}$$ that you get out of solving the orbit equation for an inverse-square potential. Getting the equation of an ellipse ($0<e<1$) is fine. You just convert from polar to cartesian coordinates to obtain the expression $$(1-e^{2})x^2+2elx+y^{2}=l^{2}$$

and then complete the square and do a bit of tidying up to get it in the form

$$\frac{(x-ea)^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$

where $a=l/(1-e^{2})$ is the semi-major axis and $b=l/\sqrt{1-e^{2}}$ is the semi-minor axis.

To get the equation of a parabolic orbit ($e=0$) is, dare I say, trivial.

I'm having a lot of trouble getting the equation of a hyperbolic orbit though, so if anyone could show me how that would be great.

Thanks!

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you could try changing $$(1-e^{2})x^2+2elx+y^{2}=l^{2}$$ to $$y^{2}-[(e^{2}-1)x^2-2elx]=l^{2}$$

then carry on from there, completing the square etc...

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    $\begingroup$ If $e>1$ then $b$ is imaginary and your ellipse equation becomes a hyperbola. $\endgroup$
    – mike stone
    Commented Jul 2, 2021 at 12:37

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