JEE Main 2023 — Parabola Question with Solution

Given,

$R$ be the focus of the parabola $y^2=20x$ and the line $y=mx+c$ intersect the parabola at two points $P$ and $Q$,

And the points $G\left(10, 10\right)$ be the centroid of the triangle $PQR$,

Now focus of the parabola $y^2=20x$ will be, $R\left(5,0\right)$

And parametric points of $PQ$ be $\left(5t^2,10t\right)$,

Now plotting the diagram we get, 

Now finding the centroid of the triangle we get,

For $x-$ coordinate we get,

$\Rightarrow \frac{5{t_1}^2+5{t_2}^2+5}{3}=10$

$\Rightarrow {t_1}^2+{t_2}^2=5 ...\left(i\right)$

Now for $y-$coordinate we get,

$\frac{10\left(t_1+t_2\right)}{3}=10$

$\Rightarrow t_1+t_2=3 ...\left(ii\right)$

Now solving both equations we get, $t_1=1, t_2=2$

So, points will be, $P\equiv (5, 10) \& Q\equiv \left(20,20\right)$

Hence, equation of $PQ=y-10=\frac{10}{15}\left(x-5\right)$

$\Rightarrow 3y-30=2x-10$

$\Rightarrow y=\frac{2}{3}x+\frac{20}{3}$, so on comparing with $y=mx+c$, we get $c-m=6$

Hence, $P{Q}^2=225+100=325$