Let P be a point on the parabola y ^ 2 = 4ax , where a > 0 . The...

Given,

Parabola ${y}^{2} = 4ax$,

Now plotting the normal to parabola at point $P\left(a{m}^{2}, - 2am\right), we get</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/a716b152-dac5-4209-b751-0649f2b39468.png" style="width: 400px; height: 219px;" /></p><p>So, equation of normal at$P\left(a{m}^{2}, - 2am\right)$is$y = mx - 2am - a{m}^{3}$</p><p>Hence, point$Q\left(2a + a{m}^{2}, 0\right)

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

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

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

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

So, possible pair from option will be,

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