Let $$P(4, -4)$$ and $$Q(9, 6)$$ be two points on the parabola, $$y^2 = 4x$$ and let X be any point on the arc POQ of this parabola, where O is the vertex of this parabola, such that the area of $$\Delta PXQ$$ is maximum. Then this maximum area (in sq. units) is:

We have the standard parabola $$y^{2}=4x$$ whose vertex is the origin $$O(0,0)$$.

A convenient way to handle points on this parabola is to use the parameter $$t$$ defined by the relations

$$x=t^{2},\qquad y=2t \qquad(\text{because }(2t)^{2}=4t^{2}=4x).$$

First we identify the parameters corresponding to the fixed points $$P(4,-4)$$ and $$Q(9,6)$$.

For $$P(4,-4):\; y=2t=-4\;\Rightarrow\; t=-2,\; x=t^{2}=(-2)^{2}=4\;(\text{checks}).$$

For $$Q(9,6):\; y=2t=6\;\Rightarrow\; t=3,\; x=t^{2}=3^{2}=9\;(\text{checks}).$$

Thus as the point $$X$$ moves along the arc $$POQ$$, its parameter $$t$$ varies in the closed interval $$[-2,3]$$ and its coordinates are

$$X\,(x,y)=\bigl(t^{2},\,2t\bigr).$$

Next we require the area of the triangle $$\Delta PXQ$$. For three points $$(x_{1},y_{1}),\,(x_{2},y_{2}),\,(x_{3},y_{3})$$ the area formula is

$$\text{Area}=\dfrac12\Bigl|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\Bigr|.$$

Taking $$(x_{1},y_{1})=(4,-4),\;(x_{2},y_{2})=(t^{2},2t),\;(x_{3},y_{3})=(9,6),$$ we substitute:

$$\begin{aligned} \text{Twice the signed area}&=4\bigl(2t-6\bigr)+t^{2}\bigl(6-(-4)\bigr)+9\bigl(-4-2t\bigr)\\[4pt] &=4(2t-6)+t^{2}(10)+9(-4-2t)\\[4pt] &=(8t-24)+10t^{2}+(-36-18t)\\[4pt] &=10t^{2}-10t-60. \end{aligned}$$

Hence the actual area is

$$A(t)=\dfrac12\,\Bigl|10t^{2}-10t-60\Bigr|=5\,\Bigl|t^{2}-t-6\Bigr|.$$

Notice that

$$t^{2}-t-6=(t-3)(t+2).$$

For $$t\in[-2,3]$$ we have $$(t+2)\ge0\text{ and }(t-3)\le0,$$ so the product is non-positive and the absolute value simply changes its sign:

$$A(t)=5\bigl(-(t-3)(t+2)\bigr)=5\,(3-t)(t+2).$$

Expanding the bracket gives

$$A(t)=5\Bigl(-t^{2}+t+6\Bigr).$$

To maximise $$A(t)$$ we consider it as a quadratic function of $$t$$. Writing

$$A(t)=5\bigl(-t^{2}+t+6\bigr)$$

the coefficient of $$t^{2}$$ is negative, so the maximum occurs at the vertex, where

$$t=-\dfrac{b}{2a}=-\dfrac{1}{2(-1)}=\dfrac12.$$

This critical value $$t=\dfrac12$$ indeed lies in the interval $$[-2,3]$$. We now compute the corresponding maximum area:

$$\begin{aligned} A_{\max}&=5\,(3-\tfrac12)\,(\tfrac12+2)\\[4pt] &=5\left(\dfrac{5}{2}\right)\left(\dfrac{5}{2}\right)\\[4pt] &=5\cdot\dfrac{25}{4}=\dfrac{125}{4}\;\text{square units}. \end{aligned}$$

Hence, the correct answer is Option C.