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 on
$\begin{array}{l}\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1\left(\text{ Suppose }\right)\\ \therefore \frac{{\mathrm{r}}^2{\cos }^2\mathrm{\theta }}{{\mathrm{a}}^2}+\frac{{\mathrm{r}}^2{\sin }^2\mathrm{\theta }}{{\mathrm{b}}^2}=1\\ {\mathrm{e}}^2{\cos }^2\mathrm{\theta }+\frac{{\mathrm{e}}^2{\sin }^2\mathrm{\theta }}{1-{\mathrm{e}}^2}=1 \left[\because {\mathrm{b}}^2={\mathrm{a}}^2\left(1-{\mathrm{e}}^2\right)\right]\\ \text{ or }{\mathrm{e}}^2{\cos }^2\mathrm{\theta }-\frac{{\mathrm{e}}^2{\cos }^2\mathrm{\theta }}{1-{\mathrm{e}}^2}=1-\frac{{\mathrm{e}}^2}{1-{\mathrm{e}}^2}\end{array}$
$$\begin{gathered} \mathrm{or} {\cos }^2\mathrm{\theta }=\frac{1}{{\mathrm{e}}^2}+\frac{1}{{\mathrm{e}}^{\mathrm{\prime }2}}-\frac{1}{{\mathrm{e}}^2{\mathrm{e}}^{\mathrm{\prime }2}} \\ \text{ or }\mathrm{\theta }={\cos }^{-1}\sqrt{\frac{1}{{\mathrm{e}}^2}+\frac{1}{{\mathrm{e}}^2}-\frac{1}{{\mathrm{e}}^2{\mathrm{e}}^{\mathrm{\prime }2}}} \\ \text{ For }\mathrm{\theta }={90}^{\circ }, \\ \begin{array}{l}\frac{{\mathrm{e}}^2+{\mathrm{e}}^2}{{\mathrm{e}}^2{\mathrm{e}}^2}=\frac{1}{{\mathrm{e}}^2{\mathrm{e}}^2}\\ \text{ or }{\mathrm{e}}^2+{{\mathrm{e}}^{\text{'}}}^2=1\end{array} \end{gathered}$$