Can it be shown that $r^2=\frac{1}{1-\varepsilon\cos(2\theta)}$ gives a conic section?

Question

I got this equation which ressembles the polar equation of a conic section. $$ r^2=\frac{1}{1-\varepsilon\cos(2\theta)} $$

I've tried to show that, for $\varepsilon>1$, this equation gives a hyperbola. When I graph it in Geogebra it looks pretty much like one, even if I change $\varepsilon$ I get an ellipse, but centered at the origin rather than at a focus point. What I can't get is a parabola, but I guess it is because of the change of origin I mentioned.

So, the thing is, can it be shown that this gives a conic section?

Answer 1

Rewrite it as $$ r^2( 1 - \epsilon \cos(2\theta)) = 1 \iff r^2 (1 - \epsilon(\cos^2 \theta - \sin^2\theta)) = 1 $$ and then expand. We have $r^2 = x^2 + y^2$, $r^2 \cos^2\theta = x^2$, and $r^2 \sin^2\theta = y^2$, so we get the traditional equation of an ellipse or a hyperbola.