JEE PYQ: Ellipse Question 12
Question 12 - 2020 (06 Sep Shift 2)
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) $e^2 + e - 1 = 0$ (c) $e^4 + e^2 - 1 = 0$ (d) $e^2 + 2e - 1 = 0$
Show Answer
Answer: (c) $e^4 + e^2 - 1 = 0$
Solution
Normal at $\left(ae, \frac{b^2}{a}\right)$: $x - ey = \frac{e(a^2 - b^2)}{a}$
$(0, -b)$ lies on it: $be = \frac{e(a^2 - b^2)}{a}$
$ab = a^2e^2 \Rightarrow \frac{b}{a} = e^2$
$1 - e^2 = e^4 \Rightarrow e^4 + e^2 - 1 = 0$
BETA
