Converting the ellipse equation $r=\frac{a\left(1-e^{2}\right)}{1+e\cos\left(\theta\right)}$ from polar to canonical cartesian form

Question

I have an ellipse with polar form equation: $$r=\frac{a\left(1-e^{2}\right)}{1+e\cos\left(\theta\right)}$$ where $e$ is eccentricity, and $a$ is semi-major axis. How do I convert this from polar form to canonical form $\frac{x^2}{b^2}+\frac{y^2}{c^2}=1?$

I tried to use $b$ = $a(1-e^2)$ and $c = 1+e \cos{\theta}$, but that didn't work, because I don't know what $\theta$ is. Is there a solution to this?

Answer 1

$$r=\frac{a\left(1-e^{2}\right)}{1+e\cos\left(\theta\right)}$$

$$r+er\cos\left(\theta\right)=a\left(1-e^{2}\right)$$

$$r^2=(a\left(1-e^{2}\right)-ex)^2$$

$$x^2+y^2=(a\left(1-e^{2}\right)-ex)^2$$

$$y^2=-(1-e^2)((x+ae)^2-a^2)$$

$$a^2(1-e^2)(\frac{(x+a e)^2}{a^2}+\frac{y^2}{a^2(1-e^2)}-1)=0.$$