## Precalculus (10th Edition)

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