If the normal at one end of a latus rectum of the ellipse $\frac{{\mathrm{x}}^2}{a^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ passes through one end of the minor axis, then $\frac{{\mathrm{e}}^4}{1-{\mathrm{e}}^2}=\mathrm{\_\_\_\_}$ (Here e is the eccentricity of the ellipse)

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

Equation of normal at $\mathrm{L}\left(\mathrm{ae}, \frac{{\mathrm{b}}^2}{\mathrm{a}}\right)$ is 

$\Rightarrow \frac{\mathrm{ax}}{\mathrm{e}}-\mathrm{ay}={\mathrm{a}}^2{\mathrm{e}}^2 \left(\therefore {\mathrm{b}}^2={\mathrm{a}}^2(1-{\mathrm{e}}^2)\right)$

since it passes through (0, -b) then 

$\begin{array}{rcl} & \Rightarrow & 0+\mathrm{ab}={\mathrm{a}}^2{\mathrm{e}}^2\\ & \Rightarrow & \mathrm{b}={\mathrm{ae}}^2\\ & \Rightarrow & {\mathrm{b}}^2={\mathrm{a}}^2{\mathrm{e}}^4\\ & \Rightarrow & {\mathrm{a}}^2(1-{\mathrm{e}}^2)={\mathrm{a}}^2{\mathrm{e}}^4\\ & \Rightarrow & \frac{{\mathrm{e}}^4}{1-{\mathrm{e}}^2}=1\end{array}$