JEE Main 2020 (Online) 3rd September Morning ShiftQuestion 1
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 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 $\frac43$, then :
Parabola
Solution
$P(3t^2,6t)$, $N(3t^2,0)$, $M(3t^2,3t)$. Line $y=3t$ meets parabola at $Q(\frac{3t^2}{4},3t)$. Slope $NQ=\frac{3t}{-9t^2/4}=-\frac{4}{3t}$; $y$-intercept $=-\frac{4}{3t}(0-3t^2)=4t=\frac43\Rightarrow t=\frac13$. $PN=6t=2$, $MQ=3t^2-\frac{3t^2}{4}=\frac{9t^2}{4}=\frac14$. Option B.