Let P be a point on the parabola y 2 = 4 a x , where a > 0 . The normal to the parabola at P meets the x…

Given,

Parabola $y^2=4ax$,

Now plotting the normal to parabola at point $P\left(a{m}^2,-2am\right)$, we get

So, equation of normal at $P\left(a{m}^2, -2am\right)$ is $y=mx-2am-a{m}^3$

Hence, point $Q\left(2a+a{m}^2,0\right)$

And focus of parabola will be $F\left(a,0\right)$

Now finding the area of $∆PFQ$ we get,

Area of $∆PFQ=\frac{1}{2}\left(a+a{m}^2\right)\times 2am=120$

$\Rightarrow a^{2}m\left(1+m^2\right)=120$

So, possible pair from option will be,

$\left(a, m\right)\equiv \left(2, 3\right)$ which satisfies above equation