Question

If the normal at one extremity of the latus rectum of ellipse passes through on extremity of the mirror axis, show that the eccentricity of the ellipse can be obtained from the equation. $e^4+e^2-1=0$.

Answer 1

Assume that the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

Equation of normal to this ellipse at $(ae,\frac{b^2}{a})$ is $\frac{x-ae}{\frac{ae}{a^2}}=\frac{y-\frac{b^2}{a}}{\frac{b^2}{a{b}^2}}$

This normal passes through $(0,-b)$ according to given condition.

$\therefore \frac{0-ae}{\frac{ae}{a^2}}=\frac{-b-\frac{b^2}{a}}{\frac{b^2}{a{b}^2}}$

$\therefore -a^2=-ab-b^2$

$\therefore -a^2=-ab-b^2$

$\therefore a^2=ab+a^2(1-e^2)$

$\therefore b=a{e}^2$

$\therefore b^2=a^2{e}^4$ (squaring both sides)

$\therefore a^2(1-e^2)=a^2{e}^4$

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

Hence proved.