Question

Change to polar coordinates the equation: $x^2=y^2(2a-x)$

Answer 1

General polar form $(r,\theta )$

$r=\sqrt{x^2+y^2}$

$x=r\cos \theta$

$y=r\sin \theta$

$\theta ={\tan }^{-1}\frac{y}{x}$

Given : $x^2=y^2(2a-x)$ ........ $\left(i\right)$

Substituting values of $x$ and $y$ in $\left(i\right),$

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

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

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