Answer
$12x^2-4y^2+64-64x=0$
Work Step by Step
Cross-multiply to obtain:
$r(2+4\cos{\theta})=8\\2r+4r\cos{\theta}=8\\2r=8-4r\cos{\theta}$
Square both sides:
$(2r)^2=\left(8-4r\cos{\theta}\right)^2\\
\\4r^2=(8-4r\cos{\theta})^2$.
We know that $r^2=x^2+y^2$ and that $x=r\cos\theta,y=r\sin\theta$.
Hence
$4r^2=(8-4r\cos{\theta})^2$ becomes:
\begin{align*}
4(x^2+y^2)&=(8-4x)^2\\
4x^2+4y^2&=64-64x+16x^2\\
0&=64-64x+16x^2-4x^2-4y^2\\
0&=12x^2-4y^2-64x+64
\end{align*}
