Question:
If the normal at an end of a latus rectum of an ellipse passes through an extermity of the minor axis, then the eccentricity e of the ellipse satisfies:
Correct Option: , 3
Solution:
Normal to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at $\left(a e, \frac{b^{2}}{a}\right)$ is
$\frac{a^{2} x}{a e}-\frac{b^{2} y}{b^{2} / a}=a^{2}-b^{2}$
$\Rightarrow x-e y=\frac{e\left(a^{2}-b^{2}\right)}{a}$...(i)
$\because(0,-b)$ lies on equation (i), then
$b e=\frac{e\left(a^{2}-b^{2}\right)}{a}$
$\Rightarrow a b=a^{2} e^{2} \Rightarrow b=a e^{2} \Rightarrow \frac{b^{2}}{a^{2}}=e^{4}$
$\therefore 1-e^{2}=e^{4} \Rightarrow e^{4}+e^{2}-1=0$