# 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 mid-point M of PN, parallel to its axis which meets the parabola at Q. If the y-intercept of the NQ is $\frac{4}{3}$, then: Option: 1 PN=4 Option: 2 Option: 3   Option: 4 PN=3

$P=(3t^{2},6t), N =(3t^{2},0)$

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

$\therefore t=\frac{1}{3}$\

$\\\therefore \mathrm{MQ}=\frac{9}{4} \mathrm{t}^{2}=\frac{1}{4} \text{ and } \mathrm{PN}=6 \mathrm{t}=2$

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