JEE PYQ: Parabola Question 7
Question 7 - 2020 (03 Sep Shift 1)
Let $P$ be a point on the parabola, $y^2 = 12x$ and $N$ be the foot of the perpendicular drawn from $P$ on the axis of the parabola. A line is now drawn through the mid-point $M$ of $PN$, parallel to its axis which meets the parabola at $Q$. If the $y$-intercept of the line $NQ$ is $\frac{4}{3}$, then:
(1) $PN = 4$
(2) $MQ = \frac{1}{3}$
(3) $MQ = \frac{1}{4}$
(4) $PN = 3$
Show Answer
Answer: (3)
Solution
$y^2 = 12x$, $a = 3$. Let $P(at^2, 2at) = (3t^2, 6t)$. $N(3t^2, 0)$, $M(3t^2, 3t)$. $QM$ is parallel to axis so $Q = \left(\frac{(3t)^2}{12}, 3t\right) = \left(\frac{3t^2}{4}, 3t\right)$. Line $NQ$ passes through $\left(0, \frac{4}{3}\right)$: solving gives $t = \frac{1}{3}$. Then $MQ = \frac{3}{4}t^2 = \frac{1}{4}$ and $PN = 2at = 2$.
BETA
