The sum of the squares of the 

perpendiculars on any tangent to the  ellipse 

$x^2/a^2+y^2/b^2=1$ from two points on the minor

 axis each at a distance  $\sqrt{a^2-b^2}$ from the centre is 

 

The eccentricity e of the given ellipse is given 

by $g^2=1-b^2/a^2\Rightarrow a^2-b^2=a^2{e}^2$  . So the point on the minor

axis, i.e. y-axis at a distance  $\sqrt{a^2-b^2}$ from the  centre $(0, 0)$

of the ellipse are (0, ± ae).  

The equation of the tangent at any point $(a \cos \theta , b \sin \theta )$

on the ellipse is $\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$ 

  So the required sum is  

$\begin{array}{l}{\left[\frac{\frac{ae\sin \theta }{b}-1}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}\right]}^2+{\left[\frac{\frac{-ae\sin \theta }{b}-1}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}\right]}^2\\ =\frac{(ae\sin \theta -b{)}^2+(ae\sin \theta +b{)}^2}{\left(b^2{\cos }^2\theta +a^2{\sin }^2\theta \right)}\times a^2\\ =\frac{2a^2\left(a^2{e}^2{\sin }^2\theta +b^2\right)}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }\end{array}$

 $=\frac{2a^2\left[\left(a^2-b^2\right){\sin }^2\theta +b^2\right)}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }=2a^2$