## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$r^2 \ \sin^2(\theta)- 2r \ \cos(\theta) =0$
The conversion of polar co-ordinates $(r, \theta)$ to rectangular coordinates $(x,y)$ can be expressed as: $x=r \ \cos(\theta)$, $y=r \ \sin(\theta)$ where, $r=\sqrt{x^2+y^2}$ We have: $y^2= 2x$ After making the substitutions above, we get: $r^2 \ \sin^2(\theta)=2 r \ \cos(\theta)$ This implies that $r^2 \ \sin^2(\theta)- 2r \ \cos(\theta) =0$ Therefore, $r^2 \ \sin^2(\theta)- 2r \ \cos(\theta) =0$