Let S and ${\mathrm{S}}^1$ are the foci of an ellipse whose eccentricity is $\frac{1}{\sqrt{2}}$ , B and ${\mathrm{B}}^1$ are the ends of minoraxis then ${\mathrm{SBS}}^1{\mathrm{B}}^1$ forms a

$$\begin{gathered} Given e=\frac{1}{\sqrt{2}} \\ Clearly \mathrm{SB}={\mathrm{S}}^1\mathrm{B}={\mathrm{S}}^1 {\mathrm{B}}^1={\mathrm{SB}}^1=\mathrm{a} \\ and {\mathrm{BB}}^1=2\mathrm{b} , {\mathrm{SS}}^1=2\mathrm{ae} \\ =\sqrt{2} \mathrm{a} \\ Now {\mathrm{b}}^2={\mathrm{a}}^2\left(1-\frac{1}{2}\right)=\frac{{\mathrm{a}}^2}{2} \\ \Rightarrow \mathrm{b}=\frac{\mathrm{a}}{\sqrt{2}} \\ Now {\mathrm{SS}}^1=\sqrt{2} \left(\sqrt{2} \mathrm{b}\right)=2\mathrm{b} \end{gathered}$$

Length of diagonals same

So,${\mathrm{SBS}}^1{\mathrm{B}}^1$ is a square