Answer
$r=\dfrac{ed}{1-e \sin \theta}$
Work Step by Step
Consider the eccentricity which is defined as $e=\dfrac{|PF|}{|Pl|}$
and $|PF|=r$ and $|Pl|=d-r \sin \alpha$
But $\alpha =2\pi-\theta$
Now, $|Pl|=d-r \sin \alpha$
This gives: $|Pl|=d-r\cos (2\pi-\theta)=d+r \sin \theta$
$\dfrac{|PF|}{|Pl|}=e \implies e=\dfrac{r}{d+r \sin \theta}$
simplify.
$r=ed+er \sin \theta \implies r(1-e \sin \theta)=ed$
Hence, $r=\dfrac{ed}{1-e \sin \theta}$
