Answer
$r^2 \ \cos^2(\theta)-4r \ \sin(\theta) =0$
Work Step by Step
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: $x^2=4y$
After making the substitutions above, we get:
$r^2 \ \cos^2(\theta)=4 r \ \sin(\theta)$
This implies that $r^2 \ \cos^2(\theta)-4r \ \sin(\theta) =0$
Therefore, $r^2 \ \cos^2(\theta)-4r \ \sin(\theta) =0$
