JEE PYQ: Ellipse Question 15
Question 15 - 2020 (08 Jan Shift 1)
Let the line $y = mx$ and the ellipse $2x^2 + y^2 = 1$ intersect at a point $P$ in the first quadrant. If the normal to this ellipse at $P$ meets the co-ordinate axes at $\left(-\frac{1}{3\sqrt{2}}, 0\right)$ and $(0, \beta)$, then $\beta$ is equal to:
(a) $\frac{2\sqrt{2}}{3}$ (b) $\frac{2}{\sqrt{3}}$ (c) $\frac{2}{3}$ (d) $\frac{\sqrt{2}}{3}$
Show Answer
Answer: (d) $\frac{\sqrt{2}}{3}$
Solution
Normal at $P$: $\frac{x}{2x_1} - \frac{y}{y_1} = \frac{1}{2}$
Passes through $\left(-\frac{1}{3\sqrt{2}}, 0\right) \Rightarrow x_1 = \frac{-1}{3\sqrt{2}}$
$y_1 = \frac{2\sqrt{2}}{3}$ (P in 1st quadrant)
$\beta = \frac{y_1}{2} = \frac{\sqrt{2}}{3}$
BETA
