1. Parametrize the parabola

For the parabola $y^2=4x,$ a standard parametric point is $P(t)=(t^2,2t).$

Given:

• $A(4,-4)$ corresponds to $t=-2$ since $( (-2)^2,2(-2))=(4,-4)$.
• $B(9,6)$ corresponds to $t=3$ since $(3^2,2\cdot 3)=(9,6)$.

Let $C=(t^2,2t), \qquad -2\le t\le 3,$ since $C$ lies on the arc $AOB$.

2. Area of triangle $ACB$

Using the determinant formula, $[ACB]=\frac12\left|x_A(y_C-y_B)+x_C(y_B-y_A)+x_B(y_A-y_C)\right|.$

Substitute: $A=(4,-4),\quad B=(9,6),\quad C=(t^2,2t).$

So, \begin{align*} 2[ACB] &=\left|4(2t-6)+t^2(6-(-4))+9((-4)-2t)\right|\ &=\left|8t-24+10t^2-36-18t\right|\ &=\left|10t^2-10t-60\right|\ &=10\left|t^2-t-6\right|\ &=10\left|(t-3)(t+2)\right|. \end{align*}

Hence, $[ACB]=5\left|(t-3)(t+2)\right|.$

Now for $-2\le t\le 3$,

• $t-3\le 0$
• $t+2\ge 0$

Therefore, $(t-3)(t+2)\le 0,$ so $\left|(t-3)(t+2)\right|=-(t-3)(t+2)=(3-t)(t+2).$

Thus, $[ACB]=5(3-t)(t+2).$

3. Maximize the area

Let $f(t)=5(3-t)(t+2).$ Expand: \begin{align*} f(t)&=5(3t+6-t^2-2t)\ &=5(-t^2+t+6)\ &=-5t^2+5t+30. \end{align*}

This is a downward-opening parabola, so its maximum occurs at the vertex: $t=\frac{-b}{2a}=\frac{-5}{2(-5)}=\frac12.$

Then \begin{align*} [ACB]_{\max} &= -5\left(\frac12\right)^2+5\left(\frac12\right)+30\ &= -5\cdot\frac14+\frac52+30\ &= -\frac54+\frac{10}{4}+30\ &= \frac54+30\ &=31\frac14. \end{align*}

4. Check options

The maximum area is $31\frac14.$ So the correct option is:

A: $31\frac14$

5. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.