Let $P$ be a point on the parabola, $y^{2}=12 x$ 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 $P N$, parallel to its axis which meets the parabola at $Q$. If the
$y$-intercept of the line $N Q$ is $\frac{4}{3}$, then :
Correct Option: , 3
$\because y^{2}=12 x$
$\therefore a=3$
Let $P\left(a t^{2}, 2 a t\right)$
$\Rightarrow N\left(a t^{2}, 0\right) \Rightarrow M\left(a t^{2}, a t\right)$
$\because$ Equation of $Q M$ is $y=a t$
So, $y^{2}=4 a x \Rightarrow x=\frac{a t^{2}}{4}$
$\Rightarrow Q\left(\frac{a t^{2}}{4}, a t\right)$
$\Rightarrow$ Equation of $Q N$ is $y=\frac{-4}{3 t}\left(x-a t^{2}\right)$
$\because Q N$ passes through $\left(0, \frac{4}{3}\right)$, then
$\frac{4}{3}=-\frac{4}{3 t}\left(-a t^{2}\right) \Rightarrow a t=1 \Rightarrow t=\frac{1}{3}$
Now, $M Q=\frac{3}{4} a t^{2}=\frac{1}{4}$ and $P N=2 a t=2$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.