I am trying to see if I am on the right track with this.
The problem: A kepler orbit (an ellipse) in Cartesian coordinates is: $$(1−\epsilon^2)x^2 + 2\alpha \epsilon x + y^2 = \alpha^2.$$ The task is to show that the major and minor axes are: $$a = \frac{\alpha}{\sqrt{1-\epsilon^2}} = \frac{k}{2|E|} \text{ and } b = \frac{\alpha}{\sqrt{1-\epsilon^2}} = \frac{L}{\sqrt{2m|E|}}$$
Well and good, I noticed that the general equation for an ellipse has that middle term $2\alpha \epsilon x $ but I could get rif of it if I just asume the ellipse's center is at the origin. Then I can say the general form for the ellipse is $\frac{x^2}{a}+\frac{y^2}{b}=1$ and go from there. When I do that I can redue the original equation to: $$\frac{(1−\epsilon^2)x^2}{\alpha^2} + \frac{y^2}{\alpha^2} = 1$$
Plugging in for a (that is, noting that under the x is the value for a) I see I can make the denominator under x equal to $\frac{\alpha^2}{1-\epsilon^2}$ which would make $a=\frac{\alpha}{\sqrt{1-\epsilon}}$ and I can do the same thing for b, getting me $b=\frac{\alpha}{\sqrt{1-\epsilon}}$ as well.
It's the next step I am a bit shaky on. Assuming $\alpha = \frac{L^2}{mk}$ I am not entirely sure how to get the last step. I was thinking that to get total energy (E) I would just add the vectors of radial and tangental velocity, and plug that into the old $KE= \frac{1}{2} mv^2$. But I am trying to determine if I am in the right ballpark. I suppose this is more physics than math per se, but the technique is heavily mathematical and that appears to be the crux of the problem.
If $\;0\le\epsilon<1$ then your equation can be written as $$(1-\epsilon^2)\left(x+\frac {\alpha\epsilon}{1-\epsilon^2}\right)^2+y^2=\alpha^2+\frac {\alpha^2\epsilon^2}{1-\epsilon^2}=\frac {\alpha^2}{1-\epsilon^2}$$or$$\left(x+\frac {\alpha\epsilon}{1-\epsilon^2}\right)^2 \cdot \frac {(1-\epsilon^2)^2}{\alpha^2}+y^2 \cdot\frac {1-\epsilon^2}{\alpha^2}=1$$It is the equation of an ellipse centered at $$\left(-\frac {\alpha\epsilon}{1-\epsilon^2},0\right)$$semimajoraxis$$a=\frac {\alpha}{1-\epsilon^2}$$semiminor axis $$b=\frac {\alpha}{\sqrt {1-\epsilon^2}}$$In fact $1-\epsilon^2<1$ so that $a^2>b^2$.
Note that $$E=KE-\frac kr=const<0$$with$$r=\sqrt{x^2+y^2}$$and$$\epsilon^2=\frac {2L^2E}{mk^2}+1$$$$a=\frac{k}{-2E}$$$$b=\frac {L}{\sqrt {-2mE}}$$