Question

The normal at the end point of a latus rectum of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ passes through the end point of its minor axis. Find the value of $e^{2}+e^{4}$ .

Answer 1

Equation of normal to ellipse at one end of latus
rectum $\left(a e , \frac{b^{2}}{a}\right)$ is
$\frac{a^{2} x}{a e}-\frac{b^{2} y}{\frac{b^{2}}{a}}=a^{2}-b^{2}$
$\Rightarrow \frac{a x}{e}-ay=a^{2}e^{2}$
Since it passes through $\left(\right.0,-b\left.\right)$ ,
$\Rightarrow 0+ab=a^{2}e^{2}$
$\Rightarrow b^{2}=a^{2}e^{4}$
$\Rightarrow a^{2}\left(1 - e^{2}\right)=a^{2}e^{4}$
$\Rightarrow 1-e^{2}=e^{4}$
$\Rightarrow e^{4}+e^{2}=1$