Answer
$r=\dfrac{ed}{1+e \sin \theta}$
Work Step by Step
Since, the eccentricity $e$ is: $e=\dfrac{|PF|}{|Pl|}$
and $|PF|=r$ ; $|Pl|=d-r \sin \theta$
Now, the equation $e=\dfrac{|PF|}{|Pl|}$ becomes:
$e= \dfrac{r}{d-r \sin \theta}$
This implies that $r=ed-er \sin \theta$
Therefore, we get, after simplification:
$r=\dfrac{ed}{1+e \sin \theta}$
