If p and p ' denote the lengths of the perpendicular from a focus and the centre of an ellipse with…

Let the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

and let $\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$  $...\left(\mathrm{i}\right)$

be a tangent to it at point $(a\cos \theta ,b\sin \theta )$. Then, $p=$ length of the perpendicular from $S(ae, 0)$ on Eq. $\left(\mathrm{i}\right)$

$p=\left|\frac{e\cos \theta -1}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}\right|$

$p^{\text{'}}=$ length of the perpendicular from $O(0,0)$ on Eq. $\left(\mathrm{i}\right)$

$\Rightarrow p^{\text{'}}=\left|\frac{1}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}\right|$

and $r=ae\cos \theta -a$

Clearly, $r{p}^{\text{'}}=ap$