There are exactly two points on the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ whose distances from its centre are the same and are equal to $\sqrt{\frac{{\mathrm{a}}^2+2{\mathrm{b}}^2}{2}}$ then the eccentricity of the ellipse is

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

Since there are exactly two points on the ellipse whose distance from the centre is same, the points would be either end points of the major axis (or) of the minor axis.

But $\frac{\sqrt{{\mathrm{a}}^2+2{\mathrm{b}}^2}}{\sqrt{2}}>\mathrm{b},$ so the points are the vertices of the major axis.

Hence $\mathrm{a}=\frac{\sqrt{{\mathrm{a}}^2+2{\mathrm{b}}^2}}{\sqrt{2}}$

squaring on both sides we get

$$\begin{gathered} \Rightarrow 2{\mathrm{a}}^2={\mathrm{a}}^2+2{\mathrm{b}}^2 \\ \Rightarrow {\mathrm{a}}^2=2{\mathrm{b}}^2 \Rightarrow \frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=\frac{1}{2} \end{gathered}$$

Now $\mathrm{e}=\sqrt{1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}$