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

Answer: 1

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