Question

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

• A. $MQ=\frac{1}{3}$
• B. $PN=3$
• C. $MQ=\frac{1}{4}$
• D. $PN=4$

Answer 1

Answer: (C)

Let $P=\left(3t^2,6t\right);N=\left(3t^2,0\right)$
$M=\left(3t^2,3t\right)$
Equation of $MQ:y=3t$
$\therefore Q=\left(\frac{3}{4}{t}^2,3t\right)$
Equation of NQ
$y=\frac{3t}{\left(\frac{3}{4}{t}^2-3t^2\right)}\left(x-3t^2\right)$
y-intercept of $NQ=4t=\frac{4}{3}\Rightarrow t=\frac{1}{3}$
$\therefore MQ=\frac{9}{4}{t}^2=\frac{1}{4}$
$PN=6t=2$