Find the eccentricity of the ellipse, if its length of the latus rectum is equal to half of its minor axis.

Let the equation of ellipse be$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Given that $\frac{2{\mathrm{b}}^2}{\mathrm{a}}=\frac{1}{2}\left(2\mathrm{b}\right)$
$\begin{array}{l}2\mathrm{b}=\mathrm{a}\Rightarrow 4{\mathrm{b}}^2={\mathrm{a}}^2\Rightarrow 4\cdot {\mathrm{a}}^2\left(1-{\mathrm{e}}^2\right)={\mathrm{a}}^2\\ \Rightarrow 1-{\mathrm{e}}^2=\frac{1}{4}\\ \Rightarrow {\mathrm{e}}^2=1-\frac{1}{4}=\frac{3}{4}\Rightarrow \mathrm{e}=\frac{\sqrt{3}}{2}\end{array}$