Let $$A(4, -4)$$ and $$B(9, 6)$$ be points on the parabola, $$y^2 = 4x$$. Let $$C$$ be chosen on the arc AOB of the parabola, where $$O$$ is the origin, such that the area of $$\triangle ACB$$ is maximum. Then, the area (in sq. units) of $$\triangle ACB$$, is:

We have the standard parabola $$y^{2}=4x$$. A convenient way to describe every point on this curve is to put $$y=2t$$. Substituting this into the equation of the parabola gives $$x=\dfrac{y^{2}}{4}=\dfrac{(2t)^{2}}{4}=t^{2}$$. Hence any point on the parabola can be written in the parametric form $$\bigl(t^{2},\,2t\bigr)$$, where $$t$$ is a real parameter.

The fixed points are $$A(4,-4)\quad\text{and}\quad B(9,6).$$ Comparing with the parametric form, $$A(4,-4)\;\Longrightarrow\;4=t^{2},\; -4=2t\;\Longrightarrow\;t=-2,$$ $$B(9,6)\;\Longrightarrow\;9=t^{2},\; 6=2t\;\Longrightarrow\;t=3.$$ Thus $$t=-2$$ corresponds to $$A$$ and $$t=3$$ corresponds to $$B$$. The point $$O(0,0)$$ corresponds to $$t=0$$. The arc $$AOB$$ therefore traces the parameter $$t$$ from $$-2$$ through $$0$$ up to $$3$$.

Let us choose a variable point $$C\bigl(t^{2},\,2t\bigr)$$ with $$-2\le t\le 3$$ on this arc. Our goal is to maximise the area of $$\triangle ACB$$.

For three points $$A(x_{1},y_{1}),\;B(x_{2},y_{2}),\;C(x_{3},y_{3})$$ the area formula is $$\text{Area}=\dfrac12\bigl|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\bigr|.$$ We first write down the coordinates explicitly:

$$\begin{aligned} A&:(x_{1},y_{1})=(4,-4),\\ B&:(x_{2},y_{2})=(9,6),\\ C&:(x_{3},y_{3})=(t^{2},2t). \end{aligned}$$ Substituting all these into the formula gives $$\begin{aligned} 2\,\text{Area}&=\bigl|\,4\,(6-2t)+9\,(2t-(-4))+t^{2}\,(-4-6)\bigr|\\[4pt] &=\bigl|\,4(6-2t)+9(2t+4)+t^{2}(-10)\bigr|. \end{aligned}$$

We now simplify each term one by one:

$$\begin{aligned} 4(6-2t)&=24-8t,\\ 9(2t+4)&=18t+36,\\ t^{2}(-10)&=-10t^{2}. \end{aligned}$$ Adding them, $$ 24-8t+18t+36-10t^{2}=60+10t-10t^{2}. $$ Hence $$ 2\,\text{Area}=|\,60+10t-10t^{2}\,|=10\,|\, -t^{2}+t+6\,|. $$

Inside the modulus we have the quadratic $$f(t)=-t^{2}+t+6.$$ Let us analyse its sign on the interval $$-2\le t\le 3$$. At the end-points,

$$\begin{aligned} f(-2)&=-(-2)^{2}+(-2)+6=-4-2+6=0,\\ f(3)&=-3^{2}+3+6=-9+3+6=0. \end{aligned}$$

The quadratic opens downwards (coefficient of $$t^{2}$$ is negative), so between the roots its value is positive. Therefore $$f(t)\ge0$$ throughout the required interval, and the modulus may be dropped:

$$2\,\text{Area}=10\bigl(-t^{2}+t+6\bigr),$$ $$\text{Area}=5\bigl(-t^{2}+t+6\bigr).$$ Call the bracketed expression $$g(t)=-t^{2}+t+6$$, so $$\text{Area}=5\,g(t).$$ To maximise the area we simply have to maximise $$g(t)$$.

The quadratic $$g(t)=-t^{2}+t+6$$ has its vertex at $$t=-\dfrac{b}{2a}=-\dfrac{1}{2(-1)}=\dfrac12,$$ using the vertex formula $$t=-\dfrac{b}{2a}$$ for $$at^{2}+bt+c$$. This value $$t=\dfrac12$$ indeed lies between $$-2$$ and $$3$$, so it gives the maximum.

Evaluating $$g(t)$$ at this point, $$\begin{aligned} g\!\left(\dfrac12\right)&=-\left(\dfrac12\right)^{2}+\dfrac12+6\\[4pt] &=-\dfrac14+\dfrac12+6\\[4pt] &=6.25. \end{aligned}$$ Finally the maximal area is $$ \text{Area}_{\max}=5\times6.25=31.25=\dfrac{125}{4}. $$

Expressed as a mixed number, $$31.25=31\dfrac14$$.

Hence, the correct answer is Option D.