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 PN, parallel to its axis which meets the parabola at Q. If the y-intercept of the line NQ is $\frac{4}{3}$ , then

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$P N = 3$

$M Q = \frac{1}{4}$

$M Q = \frac{1}{3}$

$P N = 4$

answer is C.

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Detailed Solution

We have, $y^{2} = 12 x$
Let $P (3 t^{2}, 6 t), N (3 t^{2}, 0), M (3 t^{2}, 3 t)$
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Equation of MQ is $y = 3 t$

$∴ Q = (\frac{3}{4} t^{2}, 3 t)$

Equation of NQ is $y - 0 = \frac{3 t}{(\frac{3}{4} t^{2} - 3 t^{2})} (x - 3 t^{2})$
$\Rightarrow y = \frac{3 t}{3 t^{2} (\frac{- 3}{4})} (x - 3 t^{2}) \Rightarrow y = \frac{- 4}{3 t} x + 4 t$
Given that y-intercept of NQ=4/3
$∴ 4 t = \frac{4}{3} \Rightarrow t = \frac{1}{3} ∴ M Q = 3 t^{2} - \frac{3}{4} t^{2} = \frac{9}{4} t^{2} = \frac{1}{4} a n d P N = 6 t = 2$