JEE Main 2021 — Parabola Question with Solution

We have,

$y=4x^2+1$

$L:y=x$

Let the foot of perpendicular from $P$ to line $y=x$ is $Q$.

Let $P\equiv \left(x,y\right)$, $Q\equiv \left(c,c\right)$ and $R\equiv \left(h,k\right)$ where, $R$ is the mid-point of $PQ$. 

Clearly,

$PQ\perp L$

$\Rightarrow \frac{k-c}{h-c}=-1$

$\Rightarrow c=\left(\frac{h+k}{2}\right)$

And, 

$R\equiv \left(\frac{x+c}{2},\frac{y+c}{2}\right)$

$\Rightarrow R\equiv \left(\frac{x}{2}+\frac{h}{4}+\frac{k}{4},\frac{y}{2}+\frac{h}{4}+\frac{k}{4}\right)$

Hence,

$h=\frac{x}{2}+\frac{h}{4}+\frac{k}{4}\Rightarrow x=\frac{3h}{2}-\frac{k}{2}$

$k=\frac{y}{2}+\frac{h}{4}+\frac{k}{4}\Rightarrow y=\frac{3k}{2}-\frac{h}{2}$

Now,

$y=4x^2+1$

$\Rightarrow \left(\frac{3k-h}{2}\right)=4{\left(\frac{3h-k}{2}\right)}^2+1$

$\Rightarrow \left(3k-h\right)=2{\left(3h-k\right)}^2+2$

Required locus is

$2{\left(3x-y\right)}^2+\left(x-3y\right)+2=0$