JEE Main 2023 — Parabola Question with Solution

Given parabola is

$x=4y^2$

$\Rightarrow y^2=\frac{1}{4}x$

$\Rightarrow y^2=4\times \frac{1}{16}\times x$

Equation of normal is

$y=-tx+2at+a{t}^3$

$\Rightarrow y=-tx+\frac{2}{16}t+\frac{1}{16}{t}^3$

It passes through $Q\left(0,33\right)$, so

$33=\frac{t}{8}+\frac{t^3}{16}$

$\Rightarrow t^3+2t-528=0$

$\Rightarrow \left(t-8\right)\left(t^2+8t+66\right)=0$

$\Rightarrow t=8$

So,

$P\equiv \left(h,k\right)\equiv \left(a{t}^2,2at\right)\equiv (4,1)$

Now, we have parabola 

$y^2=4\left(x+y\right)$

$\Rightarrow y^2-4y=4x$

$\Rightarrow {\left(y-2\right)}^2=4\left(x+1\right)$

Equation of directrix is

$x+1=-1$

$\Rightarrow x=-2$

So, distance of point $\left(4,1\right)$ from the line $x+2=0$ is

$=\left|\frac{4+2}{1}\right|=6$ units