# Proving distances of polar coordinates

$r\sin\theta=2, r=\frac{2}{1+\sin\theta}, 0<\theta<\pi$ Line l has the first equation, Curve c has the second. Any point on curve C has polar coordinates (a,$\phi$). The foot of the perpendicular from P onto l is N. Show that OP=PN.

Converting the polar coordinates to Cartesian gives that the coordinates of P are ($a\cos\phi,a\sin\phi$). Doing the same with the coordinates of N gives ($a\cos\phi,2$)

The distance OP is a.

The distance PN comes out as $2-a\sin\phi$

I'm not quite sure how to continue from here to prove that PN is the same as OP

Since $(a,\phi)$ lies on C, we have $a=\frac {2}{1+\sin \phi}$.
Rearranging this gives $a+a\sin \phi=2\Rightarrow 2-a\sin \phi=a$.
Hence $PN=OP$
Note. $C$ is a parabola whose focus is $O$ and directrix is $l$, and this is an expression of the geometric definition of a parabola.