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 NQ is 4 3 , then :

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Aicharya · Accepted Answer

Comparing the equation of the parabola, with standard equation $y^{2} = 4 a x$ , we get $a = 3$ .

Figure is drawn on the basis of the given information.

Parametric form of parabola is $x = a t^{2}, y = 2 a t$

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

$\Rightarrow Q = (\frac{3 t^{2}}{4}, 3 t)$

Equation of $N Q$ : $y - 0 = (\frac{3 t - 0}{\frac{3 t^{2}}{4} - 3 t^{2}}) (x - 3 t^{2})$

i.e., $y = \frac{- 4}{3 t} \cdot (x - 3 t^{2}) \Rightarrow y = \frac{- 4}{3 t} \cdot x + 4 t$

$\Rightarrow 4 t = \frac{4}{3} \Rightarrow t = \frac{1}{3}$

$P N = 6 t = 2$

$M Q = 3 t^{2} - \frac{3 t^{2}}{4} = \frac{9 t^{2}}{4} = \frac{1}{4}$

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…

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