Answer
$r=\dfrac{ed}{1-e \sin \theta}$
Work Step by Step
Since, the eccentricity $e$ is: $e=\dfrac{|PF|}{|Pl|}$
and $|PF|=r$ ; $|Pl|=d-r \sin \alpha$;
and $\alpha =2\pi-\theta$
Thus, we have $|Pl|=d-r \sin \alpha$
or, $|Pl|=d-r\cos (2\pi-\theta)$
$\implies |Pl|=d+r \sin \theta$
Now, the equation $e=\dfrac{|PF|}{|Pl|}$ gives:
$e= \dfrac{r}{d+r \sin \theta}$
This implies that $r=ed+er \sin \theta$
Therefore, we get $r=\dfrac{ed}{1-e \sin \theta}$
