JEE Main 2022 — Parabola Question with Solution

Given $P\left(a,b\right)$ is point on $y^2=8x$, such that tangent at $P$ pass through centre of $x^2+y^2-10x-14y+65=0$ i.e. $\left(5,7\right)$

Now tangent at $P\left(2t^2,4t\right)$ will be $ty=x+2t^2$

and it pass through $\left(5,7\right)$

So, $7t=5+2t^2$

$\Rightarrow t=1,t=\frac{5}{2}$

So, point $P\left(2t^2,4t\right)\Rightarrow \left(2,4\right)$ when $t=1$

and $\left(\frac{25}{2},10\right)$ when $t=\frac{5}{2}$

Now product of all possible value of $a$ will be  $A=2\times \frac{25}{2}=25$

And product of all possible value of $b$ will be $B=4\times 10=40$

$\therefore A+B=65$