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