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 $'e'$ 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
$\left(a e, \frac{b^{2}}{a}\right)$, then the equation of the normal at this end point is

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