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

Answer 1

The correct option is D $MQ=\frac{1}{4}$


Let $P=(3t^2,6t),N=(3t^2,0)$ and
$M=(\frac{3t^2+3t^2}{2},\frac{6t+0}{2})=(3t^2,3t)$

Equation of $MQ$ is $y=3t$

$Q(h,3t)$ lie on Parabola $y^2=12x$
$\Rightarrow 9t^2=12h$
$\Rightarrow h=\frac{3t^2}{4}$
$\therefore Q=(\frac{3t^2}{4},3t)$

Now, equation of $NQ$
$y-0=\frac{3t-0}{(\frac{3t^2}{4}-3t^2)}(x-3t^2)$
$\Rightarrow y=\frac{-4}{3t}(x-3t^2)$
Put $x=0$
$y=\frac{-4}{3t}(-3t^2)=4t$
$\Rightarrow 4t=\frac{4}{3}$ (Given)
$\Rightarrow t=\frac{1}{3}$
$PN=|6t|=|6\cdot \frac{1}{3}|=2$ and
$MQ=\sqrt{{(3t^2-\frac{3t^2}{4})}^2+(3t-3t{)}^2}$
$\Rightarrow MQ=\frac{9t^2}{4}=\frac{1}{4}$