# Let P be a point on the parabola,

Question:

Let $\mathrm{P}$ be a point on the parabola, $\mathrm{y}^{2}=12 \mathrm{x}$ and $\mathrm{N}$ be the foot of the perpendicular drawn from $\mathrm{P}$ on the axis of the parabola. A line is now drawn through the mid-point $\mathrm{M}$ 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. $\mathrm{MQ}=\frac{1}{3}$

2. $\mathrm{PN}=3$

3. $\mathrm{MQ}=\frac{1}{4}$

4. $\mathrm{PN}=4$

Correct Option: , 3

Solution:

Let $P=\left(3 t^{2}, 6 t\right) ; N=\left(3 t^{2}, 0\right)$

$\mathrm{M}=\left(3 \mathrm{t}^{2}, 3 \mathrm{t}\right)$

Equation of $\mathrm{MQ}: \mathrm{y}=3 \mathrm{t}$

$\therefore \quad Q=\left(\frac{3}{4} t^{2}, 3 t\right)$

$\mathrm{y}=\frac{3 \mathrm{t}}{\left(\frac{3}{4} \mathrm{t}^{2}-3 \mathrm{t}^{2}\right)}\left(\mathrm{x}-3 \mathrm{t}^{2}\right)$

$y$-intercept of $N Q=4 t=\frac{4}{3} \Rightarrow t=\frac{1}{3}$

$\therefore \quad \mathrm{MQ}=\frac{9}{4} \mathrm{t}^{2}=\frac{1}{4}$

$\mathrm{PN}=6 \mathrm{t}=2$