Let P be a point on the parabola, y2 = 12x 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 4 \over 3 , then :

Parabola PYQ — JEE Main 2020 · 3 Sep · Shift 1 · Q31 with Solution

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Correct answer: D

Given: $y^2=12x$ This is of the form $y^2=4ax$ , so $4a=12 \implies a=3$ The axis of the parabola is the $x$ -axis.

A standard parametric point on $y^2=4ax$ is: $P(at^2,2at)$ Since $a=3$ , $P=(3t^2,6t)$

The axis is the $x$ -axis, so dropping a perpendicular from $P=(3t^2,6t)$ gives $N=(3t^2,0)$

Thus, $PN=|6t|$

Midpoint of $P=(3t^2,6t)$ and $N=(3t^2,0)$ is $M=\left(3t^2,\frac{6t+0}{2}\right)=(3t^2,3t)$

Since the axis is the $x$ -axis, a line parallel to it is horizontal. So the line through $M$ is $y=3t$

This meets the parabola again at $Q$ .

Substitute $y=3t$ into $y^2=12x$ : $9t^2=12x$ $x=\frac{3t^2}{4}$ Hence, $Q=\left(\frac{3t^2}{4},3t\right)$

Points: $N=(3t^2,0),\qquad Q=\left(\frac{3t^2}{4},3t\right)$

Slope:

=\frac{3t}{-\frac{9t^2}{4}} =-\frac{4}{3t}$$ Equation through $N$: $$y=m(x-3t^2)$$ $$y=-\frac{4}{3t}(x-3t^2)$$ To find the $y$-intercept, put $x=0$: $$y=-\frac{4}{3t}(0-3t^2)=4t$$ Given that the $y$-intercept is $$\frac{4}{3}$$ so, $$4t=\frac{4}{3} \implies t=\frac{1}{3}$$ --- 7. **Now compute the required lengths** ### (i) $PN$ $$PN=|6t|=6\cdot\frac{1}{3}=2$$ So neither $PN=4$ nor $PN=3$ is correct. ### (ii) $MQ$ Points $M$ and $Q$ have the same $y$-coordinate, so $$MQ=\left|3t^2-\frac{3t^2}{4}\right| =\frac{9t^2}{4}$$ With $t=\frac13$, $$MQ=\frac{9}{4}\cdot\frac{1}{9}=\frac14$$ Thus, $$MQ=\frac14$$ --- 8. **Check options** - **A:** $MQ=\frac13$ ❌ - **B:** $PN=4$ ❌ - **C:** $PN=3$ ❌ - **D:** $MQ=\frac14$ ✅ Therefore, the correct option is: $$\boxed{D}$$