A square is inscribed inside the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ then the length of the side of the square is

Given ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1(\mathrm{a}>\mathrm{b})$

let ABCD be a square inscribed in ellipse.

length of AB=length of CD=2|k|

length of BC=length of AD=2|h|

since ABCD is a square then $2\left|\mathrm{k}\right|=2\left|\mathrm{h}\right| \Rightarrow \mathrm{h}=\mathrm{k}$

$$\begin{gathered} Since A lies on ellipse then \frac{{\mathrm{h}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{h}}^2}{{\mathrm{b}}^2}=1 \\ \Rightarrow {\mathrm{h}}^2 \left[\frac{{\mathrm{b}}^2+{\mathrm{a}}^2}{{\mathrm{a}}^2 {\mathrm{b}}^2}\right]=1 \\ \Rightarrow {\mathrm{h}}^2=\frac{{\mathrm{a}}^2{\mathrm{b}}^2}{{\mathrm{a}}^2+{\mathrm{b}}^2} \\ \Rightarrow \mathrm{h}=\frac{\mathrm{ab}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}} \end{gathered}$$

$\therefore$ length of side of square $=\frac{2\mathrm{ab}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}$