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