JEE Main 2020 — Parabola Question with Solution

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

Figure is drawn on the basis of the given information.

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

$\therefore P=\left(3t^2,6t\right)$, $N=\left(3t^2,0\right)$, $M=\left(3t^2,3t\right)$

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

Equation of $NQ$:  $y-0=\left(\frac{3t-0}{{\displaystyle \frac{3t^2}{4}-3t^2}}\right)\left(x-3t^2\right)$

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

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

$PN=6t=2$

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