Answer
$r=\dfrac{ed}{1-e \cos \theta}$
Work Step by Step
Consider the eccentricity which is defined as $e=\dfrac{|PF|}{|Pl|}$
Since, $|PF|=r$ and $|Pl|=d-r \cos \alpha$
But $\alpha =\pi-\theta$
This implies that, $|Pl|=d-r \cos (\alpha)=d-r\cos (\pi-\theta)$
$\implies |Pl|=d+r \cos \theta$
Further, we have
$\dfrac{|PF|}{|Pl|}=e \implies \dfrac{r}{d+r \cos \theta}=e $
$\implies r=ed+er \cos \theta$
Thus, $r(1-e \cos \theta)=ed \implies r=\dfrac{ed}{1-e \cos \theta}$
