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Step-by-Step Solution
Step 1: Identify the Parabola and Known Points
The given parabola is $y^2 = 4x$. Points on this parabola can be written in the parametric form as $(t^2, 2t)$. We have two specific points:
• $A(4, -4)$, which corresponds to the parameter $t = -2$ (since $t^2 = 4$ and $2t = -4$).
• $B(9, 6)$, which corresponds to the parameter $t = 3$ (since $t^2 = 9$ and $2t = 6$).
The origin $O(0,0)$ corresponds to $t=0$.
Step 2: Parametric Form for Point C
Let $C$ be any point on the arc $AOB$ of the parabola. Then we can represent $C$ by $(t^2, 2t)$, where $t$ ranges from $-2$ to $3$. We aim to find the point $C$ that maximizes the area of triangle $ACB$.
Step 3: Formula for the Area of $\triangle ACB$
The area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$ can be found using the determinant formula:
$$
\text{Area} = \frac{1}{2} \bigl|\, x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \bigr|.
$$
In our case:
• $A = (4, -4)$
• $B = (9, 6)$
• $C = (t^2, 2t)$.
Step 4: Substitute Coordinates into the Area Formula
Substituting these coordinates into the area determinant expression gives a function $A(t)$ in terms of $t$. Simplify the resulting expression to get $A(t)$.
Step 5: Find the Value of $t$ that Maximizes the Area
To maximize $A(t)$, we differentiate it with respect to $t$, set the derivative to zero, and solve for $t$. We also check boundaries $t=-2$ and $t=3$ if needed. The value of $t$ that maximizes the area lies within the interval $[-2, 3]$.
Step 6: Conclude the Maximum Area
By evaluating $A(t)$ at the critical point (and ensuring it gives a maximum), we get the maximum area of $\triangle ACB$ as:
$$
31 \tfrac{1}{4} \text{ square units}.
$$
Thus, the required maximum area is $31 \tfrac{1}{4}$.
Reference Diagram
