Answer
$4x^2+3y^2-64-16y=0$
Work Step by Step
After multiplying with the denominator of the fraction: $r(2-\sin{\theta})=8\\2r-r\sin{\theta}=8\\2r=8+r\sin{\theta}\\4r^2=(8+r\sin{\theta})^2$.
We know that $r^2=x^2+y^2$ and that $x=r\cos\theta,y=r\sin\theta$.
Hence $4r^2=(8+r\sin{\theta})^2$ becomes: $4(x^2+y^2)=(8+y)^2\\4x^2+4y^2=64+16y+y^2\\4x^2+3y^2-64-16y=0$
