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 

 

detailed solution

Correct option is A

The eccentricity e of the given ellipse is given by g2=1−b2/a2⇒a2−b2=a2e2  . So the point on the minoraxis, i.e. y-axis at a distance  a2−b2 from the  centre  (0, 0) of the ellipse are (0, ± ae).  The equation of the tangent at any point (a cos θ, b sin θ) on the ellipse is xacos⁡θ+ybsin⁡θ=1   So the required sum is  aesin⁡θb−1cos2⁡θa2+sin2⁡θb22+−aesin⁡θb−1cos2⁡θa2+sin2⁡θb22=(aesin⁡θ−b)2+(aesin⁡θ+b)2b2cos2⁡θ+a2sin2⁡θ×a2=2a2a2e2sin2⁡θ+b2b2cos2⁡θ+a2sin2⁡θ =2a2a2−b2sin2⁡θ+b2b2cos2⁡θ+a2sin2⁡θ=2a2