If the normal at an end of a latus rectum
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:

  1. (1) $e^{4}+2 e^{2}-1=0$

  2. (2) $\mathrm{e}^{2}+\mathrm{e}-1=0$

  3. (3) $e^{4}+e^{2}-1=0$

  4. (4) $\mathrm{e}^{2}+2 \mathrm{e}-1=0$


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

$b e=\frac{e\left(a^{2}-b^{2}\right)}{a}$

$\Rightarrow a b=a^{2} e^{2} \Rightarrow b=a e^{2} \Rightarrow \frac{b^{2}}{a^{2}}=e^{4}$

$\therefore 1-e^{2}=e^{4} \Rightarrow e^{4}+e^{2}-1=0$

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