Question

If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity $e$ of the ellipse satisfies :

• A. $e^4+2e^2-1=0$
• B. $e^2+2e-1=0$
• C. $e^4+e^2-1=0$
• D. $e^4+e-1=0$

Answer 1

The correct option is C $e^4+e^2-1=0$


Eqaution of normal at $(ae,\frac{b^2}{a})$
$\frac{a^{2}x}{ae}-\frac{b^{2}y}{\frac{b^2}{a}}=a^2{e}^2$
$\Rightarrow \frac{ax}{e}-ay=a^2{e}^2$
$\Rightarrow \frac{x}{e}-y=a{e}^2$
Since it passes through $(0,-b)$
$\therefore b=a{e}^2\Rightarrow b^2=a^2{e}^4$
$\Rightarrow a^2(1-e^2)=a^2{e}^4$
$\therefore e^4+e^2-1=0$