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Q.

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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a

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

b

PN = 4

c

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

d

PN = 3

answer is A.

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Detailed Solution

$M = (3 t^{2}, 3 t)$

Question Image

MQ line is y=3t

$Solving y^{2} = 12 x, y = 3 t$

$\begin{array}{l} \Rightarrow x = \frac{3 t^{2}}{4} \\ Q = (\frac{3 t^{2}}{4}, 3 t), N = (3 t^{2}, 0) \\ Equation of NQ is 4 x + 3 t y = 12 t^{2} \\ x = 0 \Rightarrow y = 4 t = \frac{4}{3} (given) \\ \Rightarrow t = \frac{1}{3} \\ M Q = \frac{1}{4} \end{array}$

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