Two concentric ellipses are such that the foci of one are on the other and their major axes are equal. Let e and e' be their eccentricities. Then,

Clearly, O is the midpoint of SS', and HH'.Therefore, the diagonals of quadrilateral HSH'S' bisect each other. So it is a parallelogram.Let H'OH = 2r. Then, OH = r = ae'.The point H lies onx2a2+y2b2=1( Suppose )∴r2cos2⁡θa2+r2sin2⁡θb2=1e2cos2⁡θ+e2sin2⁡θ1−e2=1    ∵b2=a21−e2 or e2cos2⁡θ−e2cos2⁡θ1−e2=1−e21−e2or cos2⁡θ=1e2+1e′2−1e2e′2  or θ=cos−1⁡1e2+1e2−1e2e′2  For θ=90∘, e2+e2e2e2=1e2e2 or e2+e'2=1