The sum of the squares of the perpendiculars on any tangent to the ellipse $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$   from two points on the minor axis each at a distance $$\sqrt {{a^2} - {b^2}} $$   from the centre is :

Solution :
$$\sqrt {{a^2} - {b^2}} = \pm ae\left( {0,\,ae} \right)$$
So, the points are $$\left( {ae,\,0} \right)$$  and $$\left( {0,\, - ae} \right)$$
Let $$\frac{x}{a}\cos \,\theta + \frac{y}{b}\sin \,\theta = 1$$     be a tangent then sum of squares of perpendicular from these points is
$$\eqalign{ & = \frac{{{{\left( {1 - \frac{{ae}}{b}\sin \,\theta } \right)}^2} + {{\left( {1 + \frac{{ae}}{b}\sin \,\theta } \right)}^2}}}{{\frac{{{{\cos }^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\theta }}{{{b^2}}}}} \cr & = \frac{{2\left( {1 + \frac{{{a^2}{e^2}}}{{{b^2}}}{{\sin }^2}\theta } \right)}}{{\frac{{{{\cos }^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\theta }}{{{b^2}}}}} \cr & = 2{a^2}\left( {\frac{{{b^2} + \left( {{a^2} - {b^2}} \right){{\sin }^2}\theta }}{{{b^2}\left( {1 - \sin \,\theta } \right) + {a^2}{{\sin }^2}\theta }}} \right) \cr & = 2{a^2} \cr} $$