Consider the equation for a conic in polar co-ordinates $(r,\theta)$
$$r = \frac{k}{1 - e\cos(\theta)} \qquad \qquad (1)$$
in the case where $k > 0$ and $e > 1$. Show that equation $(1)$ can be re-written in cartesian coordinates $(x, y)$ as $$ \frac{(x-x_0)^2}{a^2} - \frac{y^2}{b^2} = 1 \qquad \qquad (2) $$
where $a^2$, $b^2$ and $x_0$ are functions of $k$ and $e$. What type of conic does equation $(2)$ represent?
Here is my attempt.
Take $$x = r\cos(\theta)$$ and $$y = r\sin(\theta)$$. Now $r - re\cos(\theta) = k \Longleftrightarrow r = k + r\cos(\theta)$.
Squaring leads to $$r^2 = k^2 + r^2 e^2 \cos^2(\theta) + 2kre\cos(\theta)$$ which implies $$x^2 + y^2 = k^2 +e^2 x^2 + 2kex$$
Since $e>1$ then $e$ does not equal $1$ hence $1-e^2$ does not equal $0$ so we can complete the square for the equation
$$y^2 = k^2 + (e^2 - 1)x^2 + 2kex$$
But I'm not sure what i have to do from here on.