Question

Change to polar form : $x^2=y^2(2a-x)$

Answer 1

$x^2=y^2(2a-x)$

$x=r\cos \theta$ & $y=r\sin \theta$

$r^2{\cos }^2\theta =r^2{\sin }^2\theta (2a-r\cos \theta )$

$r^2{\cos }^2\theta =2a{r}^2{\sin }^2\theta -r^3{\sin }^2\theta \cos \theta$

$r^2{\cos }^2\theta -2a{r}^2{\sin }^2\theta +r^3{\sin }^2\theta \cos \theta =0$.