If p and p^ ' denote the lengths of the perpendicular from a focus...

Let the ellipse be $\dfrac{x2}{a2} + \dfrac{y2}{b2} = 1$

and let $\dfrac{xcos\theta}{a} + \dfrac{ysin\theta}{b} = 1... ( i )$

be a tangent to it at point $( acos\theta, bsin\theta )$. Then, $p =$ length of the perpendicular from $S ( ae, 0 )$ on Eq. $( i )$

$p = \left|\dfrac{ecos\theta - 1}{\dfrac{\cos^{2}\theta}{a2} + \dfrac{\sin^{2}\theta}{b2}}\right|$

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

$⇒ p^{'} = \left|\dfrac{1}{\dfrac{\cos^{2}\theta}{a2} + \dfrac{\sin^{2}\theta}{b2}}\right|$

and $r = aecos\theta - a$

Clearly, $rp^{'} = ap$