Let P 4 , - 4 and Q 9 , 6 be two points on the parabola, y 2 = 4 x and let X be any point on the arc P O Q…

Two different approaches we can use here.

Approach 1:

Let $X$ be $\left(t^2, 2t\right)$, then

Area of $\mathrm{\Delta }PXQ=\frac{1}{2}\left|\begin{array}{ccc}{t}^2& 2t& 1\\ 9& 6& 1\\ 4& -4& 1\end{array}\right|$

$\Delta =\frac{1}{2}.10\left|t^2-t-6\right| ...\left(\mathrm{i}\right)$

For maxima, differentiating both the sides with respect to $t$ and equating it to  zero, we get

${\Delta }^{\prime }=0\Rightarrow 5\left(2t-1\right)=0\Rightarrow t=\frac{1}{2}$

Hence, area of $\Delta PXQ=5\left|{\left(\frac{1}{2}\right)}^2-\frac{1}{2}-6\right|=5\left|\frac{1}{4}-\frac{1}{2}-6\right|=5\left|\frac{1-2-24}{4}\right|=\frac{125}{4}$ sq.units (using equation $\left(\mathrm{i}\right)$)

Approach 2:

For maximum area tangent to the parabola at $X$ must be parallel to $PQ$. Let $X\left(t^2, 2t\right)$, then

$2y\frac{dy}{dx}=4\Rightarrow \frac{dy}{dx}=\frac{2}{y}\Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(t^2,2t\right)}=\frac{1}{t}$

Also, slope of line $PQ=\frac{6+4}{9-4}=2$$$

Since, both the slopes are equal.

Thus, $t=\frac{1}{2}\Rightarrow X\left(\frac{1}{4}, 1\right)$

Therefore, area of $\Delta PXQ=\frac{1}{2}\left|\begin{array}{ccc}4& -4& 1\\ \frac{1}{4}& 1& 1\\ 9& 6& 1\end{array}\right|=\frac{1}{2}\left|4\left(1-6\right)-\left(-4\right)\left(\frac{1}{4}-9\right)+1\left(\frac{3}{2}-9\right)\right|=\frac{1}{2}\left|-20-35-\frac{15}{2}\right|=\frac{125}{4}$ sq.units.