#### 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 $|Pl|=d-r \sin \theta$
Thus, we have
$e=\dfrac{|PF|}{|Pl|}=e \implies e=\dfrac{r}{d-r \sin (\theta)}$
After simplifications, we get
$r=ed-er \sin (\theta)$
$\implies r(1+e \sin \theta)=ed$
Thus, $r=\dfrac{ed}{1+e \sin \theta}$