I was working on this and I wanted to be sure I wasn't too far off.
Given: $\frac{\alpha}{r} = 1 + \epsilon \cos \theta$ where $\epsilon$ is eccentricity.
Also $\frac{(x + x_0)^2}{A^2} = \frac{y^2}{B^2} = 1$ -- this is a hyperbola
So we want to find $x_0$, A, and B in terms of $\alpha$ and $\epsilon$.
So I convert to cartesian coordinates thus: $\alpha = r + r \epsilon \cos \theta$ and I get: $\alpha = \sqrt{x^2 + y^2} + \epsilon \sqrt {x^2+y^2}\frac {x}{\sqrt{x^2+y^2}}=\sqrt{x^2 + y^2}+\epsilon x$
And this can be rearranged also so that $\sqrt{x^2+y^2} = \alpha - \epsilon x$
So far so good. We can square both sides of $\alpha = \sqrt{x^2 + y^2}+\epsilon x$ to get: $\alpha^2 = x^2 + y^2+2\epsilon x\sqrt{x^2 + y^2}+\epsilon^2x^2$
$\alpha^2 = x^2 + y^2+2\epsilon x(\alpha - \epsilon x)+\epsilon^2x^2 = x^2 + y^2+2\alpha \epsilon x - 2\epsilon^2 x^2+\epsilon^2x^2 = x^2 + y^2+2\alpha \epsilon x - \epsilon^2 x^2$
I want to get tis into the form $\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1$. And it's here that I feel like I am losing the plot. There's a transformation I can do with ellipses but I wasn't sure if it would work here.
Am I doing anything wildly wrong? thanks, I know this should likely be easy enough.
If $\;\epsilon>1$ then your equation can be written as $$\left(x-\frac {\alpha\epsilon}{\epsilon^2-1}\right)^2 \cdot \frac {(\epsilon^2-1)^2}{\alpha^2}-y^2 \cdot\frac {\epsilon^2-1}{\alpha^2}=1$$It is the equation of a hyperbola centered at $$\left(\frac {\alpha\epsilon}{\epsilon^2-1},0\right)$$with$$a^2=\frac {\alpha^2}{(\epsilon^2-1)^2}$$and$$b^2=\frac {\alpha^2}{\epsilon^2-1}$$
See my answer.