Answer
$-48x^2+16y^2-64+128x=0$
Work Step by Step
After multiplying with the denominator of the fraction: $r(4+8\cos{\theta})=8\\4r+8r\cos{\theta}=8\\4r=8-8r\cos{\theta}\\16r^2=(8-8r\cos{\theta})^2$.
We know that $r^2=x^2+y^2$ and that $x=r\cos\theta,y=r\sin\theta$.
Hence $16r^2=(8-8r\cos{\theta})^2$ becomes: $16(x^2+y^2)=(8-8x)^2\\16x^2+16y^2=64-128x+64x^2\\-48x^2+16y^2-64+128x=0$
