If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity e of the ellipse satisfies:

Normal to the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \mathrm{at} \left(\mathrm{ae},\frac{{\mathrm{b}}^2}{\mathrm{a}}\right)$ is 
$\begin{array}{l}\frac{{\mathrm{a}}^2\mathrm{x}}{\mathrm{ae}}-\frac{{\mathrm{b}}^2\mathrm{y}}{{\mathrm{b}}^2/\mathrm{a}}={\mathrm{a}}^2-{\mathrm{b}}^2\\ \Rightarrow \mathrm{x}-\mathrm{ey}=\frac{\mathrm{e}\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)}{\mathrm{a}} .....\left(\mathrm{i}\right)\end{array}$
$\because$ (0,-b) lies on equation (i), then $\mathrm{be}=\frac{\mathrm{e}\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)}{\mathrm{a}}$
$\begin{array}{l}\Rightarrow \mathrm{ab}={\mathrm{a}}^2{\mathrm{e}}^2\Rightarrow \mathrm{b}={\mathrm{ae}}^2\Rightarrow \frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}={\mathrm{e}}^4\\ \therefore 1-{\mathrm{e}}^2={\mathrm{e}}^4\Rightarrow {\mathrm{e}}^4+{\mathrm{e}}^2-1=0\end{array}$