If two points are taken on the minor axis of an ellipse at the same distance from the centre as the foci, then find for which the sum of the squares of the perpendiculars from these points on any tangent to the ellipse is ^2 .

Step-by-step solution

Let the equation of the ellipse be x^2 a^2 + y^2 b^2 =1 Then, the distance of a focus from the centre =a e=a 1-b^2 / a^2 = a^2-b^2 , so that the two points of minor axis are S ₁ (0, a ^2- b ^2 ) and S ₁ ^ (0,- a ^2- b ^2 ) . Now any tangent to the ellipse is y = mx + a ^2 ~m ^2+ b ^2 where m is a parameter. The sum of the squares of the perpendiculars on this tangent from the two points S₁ and S₁ ^ is aligned & ( a^2-b^2 - a^2 m^2+b^2 1+ m ^2 )^2+ ( - a^2-b^2 - a^2 m^2+b^2 1+ m ^2 )^2 & =2 (a^2-b^2+a^2 m^2+b^2 ) /