Question

If the normal drawn at one end of the latus rectum of the ellipse $b^2{x}^2+a^2{y}^2=a^2{b}^2$ with eccentricity ${}^{\prime }{e}^{\prime }$ passes through one end of the minor axis. Then,

• A. $e^4+e^2=2$
• B. $e^4-e^2=1$
• C. $e^4+e^2=1$
• D. $e^2+e=1$

Answer 1

Answer: (C)

Given, equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,let
$(a>b)$ the an end point of a lat us rectum be
$\left(ae,\frac{b^2}{a}\right)$, then the equation of the normal at this end point is

$\frac{x-ae}{\frac{ae}{a^2}}=\frac{y-\frac{b^2}{a}}{\frac{b^2}{a{b}^2}}$
It will pass through the end of the minor axis
$(0,-b)$, if $-a^2=-ab-b^2$
$\Rightarrow 1=\frac{b}{a}+{\left(\frac{b}{a}\right)}^2\Rightarrow 1-{\left(\frac{b}{a}\right)}^2=\frac{b}{a}$
$\Rightarrow e^2=\sqrt{1-e^2}\left[\because e=\sqrt{1-{\left(\frac{b}{a}\right)}^2}\right]$
$\Rightarrow e^4+e^2=1$