If the distance between the foci of an ellipse is half the length of its latus rectum, then the eccentricity of the ellipse is

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

Given that, distance beteeen  foci=length of latus rectum.

$$\begin{gathered} \Rightarrow 2\mathrm{ae}=\frac{1}{2}\left(2\frac{{\mathrm{b}}^2}{\mathrm{a}}\right) \\ \Rightarrow 2{\mathrm{a}}^2\mathrm{e}={\mathrm{b}}^2 \\ \Rightarrow 2{\mathrm{a}}^2\mathrm{e}={\mathrm{a}}^2(1-{\mathrm{e}}^2) \\ \Rightarrow {\mathrm{e}}^2+2\mathrm{e}-1=0 \\ \Rightarrow \mathrm{e}=\frac{-2\pm \sqrt{4+4}}{2} \\ \Rightarrow \mathrm{e}=\frac{-2\pm 2\sqrt{2}}{2} \\ \Rightarrow \mathrm{e}=\sqrt{2}-1 (\because 0<\mathrm{e}<1) \end{gathered}$$