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

Options

A. `31(3)/(4)`
B. `32`
C. `30(1)/(2)`
D. `31(1)/(4)`

Answer

To find the maximum area of triangle ACB where A(4, -4), B(9, 6), and C is a point on the parabola \(y^2 = 4x\), we will follow these steps:

### Step 1: Understand the parabola and points
The equation of the parabola is given by \(y^2 = 4x\). The points A and B are provided as:
- \(A(4, -4)\)
- \(B(9, 6)\)

### Step 2: Determine the coordinates of point C
Since point C lies on the parabola, we can express its coordinates as \(C(x, y)\). From the parabola's equation, we have:
\[ y = 2\sqrt{x} \]
Thus, the coordinates of point C can be written as:
\[ C(x, 2\sqrt{x}) \]

### Step 3: Use the formula for the area of triangle ACB
The area \(A\) of triangle ACB can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates of points A, B, and C:
- \(A(4, -4)\) → \(x_1 = 4, y_1 = -4\)
- \(B(9, 6)\) → \(x_2 = 9, y_2 = 6\)
- \(C(x, 2\sqrt{x})\) → \(x_3 = x, y_3 = 2\sqrt{x}\)

The area becomes:
\[ A = \frac{1}{2} \left| 4(6 - 2\sqrt{x}) + 9(2\sqrt{x} + 4) + x(-4 - 6) \right| \]

### Step 4: Simplify the area expression
Now, simplifying the expression:
\[ A = \frac{1}{2} \left| 24 - 8\sqrt{x} + 18\sqrt{x} + 36 - 10x \right| \]
Combining like terms:
\[ A = \frac{1}{2} \left| 60 + 10\sqrt{x} - 10x \right| \]
This simplifies to:
\[ A = 5 \left| 6 + \sqrt{x} - x \right| \]

### Step 5: Maximize the area
To maximize the area, we need to differentiate \(A\) with respect to \(x\) and set the derivative to zero:
\[ \frac{dA}{dx} = 5 \left( \frac{1}{2\sqrt{x}} - 1 \right) = 0 \]
This gives:
\[ \frac{1}{2\sqrt{x}} = 1 \implies \sqrt{x} = \frac{1}{2} \implies x = \frac{1}{4} \]

### Step 6: Calculate the maximum area
Substituting \(x = \frac{1}{4}\) back into the area formula:
\[ A = 5 \left( 6 + \frac{1}{2} - \frac{1}{4} \right) = 5 \left( 6 + 0.5 - 0.25 \right) = 5 \left( 6.25 \right) = 31.25 \]

### Step 7: Conclusion
Thus, the maximum area of triangle ACB is:
\[ \text{Area} = 31 \frac{1}{4} \text{ square units} \]