Question

If the normal at one end of the latus rectum of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through one end of the minor axis, then the value of $e^4+e^2$ is, where 'e' is the eccentricity

• A. 4
• B. 1
• C. 2

Answer 1

The correct option is B 1

Normal at extremity of latus rectum is $x-ey=a{e}^3$
It passes through $(0,\pm b)$
$\Rightarrow \pm b=a{e}^2\Rightarrow b^2=a^2{e}^4\Rightarrow 1-e^2=e^4$