JEE PYQ: Ellipse Question 8
Question 8 - 2020 (04 Sep Shift 2)
Let $x = 4$ be a directrix to an ellipse whose centre is at the origin and its eccentricity is $\frac{1}{2}$. If $P(1, \beta)$, $\beta > 0$ is a point on this ellipse, then the equation of the normal to it at $P$ is:
(a) $4x - 3y = 2$ (b) $8x - 2y = 5$ (c) $7x - 4y = 1$ (d) $4x - 2y = 1$
Show Answer
Answer: (d) $4x - 2y = 1$
Solution
$\frac{a}{e} = 4 \Rightarrow a = 2$
$b^2 = 4(1 - \frac{1}{4}) = 3$
Ellipse: $\frac{x^2}{4} + \frac{y^2}{3} = 1 \Rightarrow 3x^2 + 4y^2 = 12$
$P(1, \beta)$: $3 + 4\beta^2 = 12 \Rightarrow \beta = \frac{3}{2}$
Normal at $P(1, \frac{3}{2})$: $4x - 2y = 1$
BETA
