If two points are taken on the minor axis of the ellipse x^ 2 25 +...

Given ellipse is& $\dfrac{x^{2}}{25} + \dfrac{y^{2}}{9} = 1$ … \left(1\right)$<br />$e = \sqrt{1 - \dfrac{9}{25}} = \dfrac{4}{5}$~<br />$S\left(0,5\left(\dfrac{4}{5}\right)\right) = S ( 0,4 )$<br />$S^{'}\left(0, - 5\left(\dfrac{4}{5}\right)\right) = S^{'} ( 0, - 4 )$<br /><img src="/api/proxy-image?url=https$y = mx + \sqrt{25m^{2} + 9}$where,~$m$is a parameter.</p><p>Now, the sum of the squares of$⊥~distances on this tangent from the points&S and S^{'} is
$= \left(\dfrac{m\left(0\right) - 4 + \sqrt{25m^{2} + 9}}{m^{2} + 1}\right)2 + \left(\dfrac{m\left(0\right) + 4 + \sqrt{25m^{2} + 9}}{m^{2} + 1}\right)2$
$= \dfrac{2}{( m^{2} + 1 )}\left\{{4}^{2} + ( 25m^{2} + 9 )\right\} = 50$