Let A(4, - 4) and B(9, 6) be points on the parabola, y ^ 2 = 4x ....

Let the coordinates of $C$ be \left({t}^{2}, 2t\right)$.<br /><br />Area of$∆ ABC = \dfrac{1}{2}\left|$\begin{cases}{t}^{2} & 2t & 1 \\ 9 & 6 & 1 \\ 4 & - 4 & 1\end{cases}$\right|$<br /><br /><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/fxxbaheoukaahekknvirjtznwxhovcyerjqkiutjvxiatyvuff.png" /><br /><br />$∆ = 5\left|\left(6 + t - {t}^{2}\right)\right|, $t \in \left(- 2, 3\right)$

$\dfrac{d ∆}{dt} = 1 - 2t = 0 ⇒ t = \dfrac{1}{2}$

$\dfrac{{d}^{2} △}{d{t}^{2}} = - 2$, hence area is maximum at $t = \dfrac{1}{2}$

∴ {\Delta}_\max = \dfrac{125}{4} = 31\dfrac{1}{4}