If the normal at an end of a latus rectum

Normal to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at $\left(a e, \frac{b^{2}}{a}\right)$ is

$\frac{a^{2} x}{a e}-\frac{b^{2} y}{b^{2} / a}=a^{2}-b^{2}$

$\Rightarrow x-e y=\frac{e\left(a^{2}-b^{2}\right)}{a}$...(i)

$\because(0,-b)$ lies on equation (i), then

$b e=\frac{e\left(a^{2}-b^{2}\right)}{a}$

$\Rightarrow a b=a^{2} e^{2} \Rightarrow b=a e^{2} \Rightarrow \frac{b^{2}}{a^{2}}=e^{4}$

$\therefore 1-e^{2}=e^{4} \Rightarrow e^{4}+e^{2}-1=0$