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