Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.3 Polar Coordinates - Exercises - Page 617: 16

$$3x^2 -2x +4y^2=1.$$

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 $$3x^2 -2x +4y^2=1.$$