## Calculus (3rd Edition)

$$3x^2 -2x +4y^2=1.$$
We have $$r=\frac{1}{2-\cos\theta}\Longrightarrow 2r-r\cos\theta =1.$$ Recall that $r\cos \theta=x$ and $r^2=x^2+y^2$: Hence, we get $$2\sqrt{x^2+y^2}-x =1\Longrightarrow 4(x^2+y^2)=(1+x)^2=1+2x+x^2.$$ Finally, we get $$3x^2 -2x +4y^2=1.$$