This problem requires a blend of coordinate geometry, parametric equations, and differential calculus to find the maximum area of a triangle with one vertex moving along a parabolic arc.

Let's begin by outlining the fundamental concepts and the strategy we'll employ.

Key Concepts and Strategy

1. Parametric Representation of a Parabola: For a parabola of the form $y^2 = 4ax$, any point on the parabola can be conveniently expressed in parametric coordinates as $(at^2, 2at)$. This representation simplifies calculations significantly as it reduces two variables ($x, y$) to a single parameter ($t$).
2. Area of a Triangle using Coordinates: Given three vertices $P_1(x_1, y_1)$, $P_2(x_2, y_2)$, and $P_3(x_3, y_3)$, the area of the triangle formed by these points can be calculated using the determinant formula: $\text{Area} = \frac{1}{2} \left| \det \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{pmatrix} \right|$ This formula is robust and efficient, especially when dealing with variable coordinates. The absolute value ensures the area is always positive.
3. Maximization using Differential Calculus: To find the maximum value of a function, we use its first derivative.
   • We set the first derivative of the area function (with respect to the parameter $t$) to zero to find critical points.
   • The second derivative test (or analyzing the nature of the function) helps confirm if a critical point corresponds to a maximum.
4. Geometric Condition for Maximum Area: For a triangle with a fixed base (say, AB), its area is maximized when the height from the third vertex (C) to the base is maximum. This geometrically implies that the tangent to the curve at point C must be parallel to the base AB. This condition provides an elegant alternative or verification method for finding the optimal position of C.

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Step-by-Step Solution

1. Parameterizing the Parabola and Identifying Points

The given parabola is $y^2 = 4x$.

• Why this step: Using a parametric form simplifies the coordinates of point C, making the area calculation a function of a single variable $t$.
• Comparing $y^2 = 4x$ with the standard form $y^2 = 4ax$, we see that $a=1$.
• Therefore, any point on this parabola can be represented parametrically as $P(t^2, 2t)$.

Let's find the parameter values for the given fixed points A and B:

• For point $A(4, -4)$: We use the $y$-coordinate to find $t_A$: $2t_A = -4 \implies t_A = -2$. (We can verify with the $x$-coordinate: $x_A = t_A^2 = (-2)^2 = 4$, which matches.)
• For point $B(9, 6)$: We use the $y$-coordinate to find $t_B$: $2t_B = 6 \implies t_B = 3$. (We can verify with the $x$-coordinate: $x_B = t_B^2 = (3)^2 = 9$, which matches.)

Point $C$ is chosen on the arc $AOB$ of the parabola, where $O$ is the origin. The origin $(0,0)$ corresponds to $t=0$ (since $0^2=0$ and $2(0)=0$). This means that the parameter $t$ for point $C$ must lie between $t_A = -2$ and $t_B = 3$. So, for point $C(t^2, 2t)$, we have $t \in (-2, 3)$. This range is crucial for ensuring we are considering the correct arc and for interpreting the sign of the area function later.

2. Expressing the Area of $\triangle ACB$ as a Function of $t$

• Why this step: Our goal is to maximize the area, so we need a mathematical expression for the area in terms of the variable parameter $t$.

• Let the coordinates of the vertices be $A(x_1, y_1) = (4, -4)$, $B(x_2, y_2) = (9, 6)$, and $C(x_3, y_3) = (t^2, 2t)$.

• The area of $\triangle ACB$, denoted by $\mathcal{A}(t)$, is given by the determinant formula: $\mathcal{A}(t) = \frac{1}{2} \left| \det \begin{pmatrix} 4 & -4 & 1 \\ 9 & 6 & 1 \\ t^2 & 2t & 1 \end{pmatrix} \right|$

• Let's expand the determinant along the third column for clarity. Remember the sign pattern for cofactor expansion: $(+ - +)$. $D = 1 \cdot (9 \cdot 2t - 6 \cdot t^2) - 1 \cdot (4 \cdot 2t - (-4) \cdot t^2) + 1 \cdot (4 \cdot 6 - (-4) \cdot 9)$ $D = (18t - 6t^2) - (8t + 4t^2) + (24 + 36)$

• Now, simplify the expression: $D = 18t - 6t^2 - 8t - 4t^2 + 60$ $D = -10t^2 + 10t + 60$

• Why check the sign of D: The area formula uses an absolute value. To remove it correctly, we need to know if $D$ is positive or negative within the relevant range of $t$.

• Consider the quadratic function $f(t) = -10t^2 + 10t + 60$. To find its roots, set $f(t)=0$: $-10t^2 + 10t + 60 = 0$ Divide by $-10$: $t^2 - t - 6 = 0$ Factor the quadratic: $(t-3)(t+2) = 0$ The roots are $t=3$ and $t=-2$.

• Since the coefficient of $t^2$ in $f(t)$ is negative ($-10$), the parabola represented by $f(t)$ opens downwards. This means that for any $t$ value strictly between its roots, $f(t)$ will be positive.

• Our range for $t$ is $t \in (-2, 3)$, which is exactly between the roots. Therefore, for $t \in (-2, 3)$, $D$ is positive, and we can remove the absolute value signs: $\mathcal{A}(t) = \frac{1}{2} (-10t^2 + 10t + 60)$ $\mathcal{A}(t) = -5t^2 + 5t + 30$ This is our area function, which we need to maximize.

3. Maximizing the Area Function

• Why this step: We use calculus to find the value of $t$ that yields the largest possible area.

• To find the maximum area, we take the first derivative of $\mathcal{A}(t)$ with respect to $t$ and set it to zero to find the critical points. $\frac{d\mathcal{A}}{dt} = \frac{d}{dt}(-5t^2 + 5t + 30)$ $\frac{d\mathcal{A}}{dt} = -10t + 5$

• Set the derivative to zero: $-10t + 5 = 0$ $10t = 5$ $t = \frac{5}{10} = \frac{1}{2}$

• This value $t = \frac{1}{2}$ is indeed within our allowed range for point C, $t \in (-2, 3)$.

• Why confirm it's a maximum: We use the second derivative test to ensure this critical point corresponds to a maximum, not a minimum or an inflection point. $\frac{d^2\mathcal{A}}{dt^2} = \frac{d}{dt}(-10t + 5)$ $\frac{d^2\mathcal{A}}{dt^2} = -10$ Since the second derivative is $-10$, which is negative ($< 0$), the function $\mathcal{A}(t)$ has a local maximum at $t = \frac{1}{2}$.

Alternative (Geometric) Verification:

• Why this alternative: As discussed in the key concepts, the area of $\triangle ACB$ is maximized when the tangent to the parabola at C is parallel to the chord AB. This provides an independent way to find the optimal $t$ and verify our calculus result.

1. Slope of chord AB: Using points $A(4, -4)$ and $B(9, 6)$: $m_{AB} = \frac{y_B - y_A}{x_B - x_A} = \frac{6 - (-4)}{9 - 4} = \frac{10}{5} = 2$
2. Slope of tangent at C: The equation of the parabola is $y^2 = 4x$. Differentiate implicitly with respect to $x$: $2y \frac{dy}{dx} = 4$ $\frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y}$ At point $C(t^2, 2t)$, the slope of the tangent is $m_{tangent} = \frac{2}{2t} = \frac{1}{t}$.
3. Equating slopes for maximum area: For maximum area, $m_{tangent} = m_{AB}$: $\frac{1}{t} = 2$ $t = \frac{1}{2}$ This confirms our result obtained through differential calculus, significantly increasing our confidence in the solution.

4. Calculating the Maximum Area

• **Why