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

Answer

To solve the problem, we need to find the maximum area of triangle ACB, where points A(4, -4), B(9, 6), and C(x, y) lie on the parabola defined by \(y^2 = 4x\).

### Step-by-Step Solution:

1. **Identify Coordinates of Points A and B**:
- Point A is given as \(A(4, -4)\).
- Point B is given as \(B(9, 6)\).

2. **Determine the Coordinates of Point C**:
- Since point C lies on the parabola \(y^2 = 4x\), we can express the coordinates of point C as \(C(x, y)\).
- From the parabola equation, we have \(y = 2\sqrt{x}\) (taking the positive root since we are looking for the area and both points A and B have positive y-coordinates).

3. **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 \(A(4, -4)\), \(B(9, 6)\), and \(C(x, 2\sqrt{x})\):
\[ \text{Area} = \frac{1}{2} \left| 4(6 - 2\sqrt{x}) + 9(2\sqrt{x} + 4) + x(-4 - 6) \right| \]
- Simplifying this:
\[ = \frac{1}{2} \left| 24 - 8\sqrt{x} + 18\sqrt{x} + 36 - 10x \right| \]
\[ = \frac{1}{2} \left| 60 + 10\sqrt{x} - 10x \right| \]
\[ = 30 + 5\sqrt{x} - 5x \]

4. **Maximizing the Area**:
- To find the maximum area, we differentiate the area function with respect to \(x\):
\[ \frac{dA}{dx} = \frac{5}{2\sqrt{x}} - 5 \]
- Setting the derivative equal to zero for maximization:
\[ \frac{5}{2\sqrt{x}} - 5 = 0 \]
\[ \frac{5}{2\sqrt{x}} = 5 \]
\[ 2\sqrt{x} = 1 \quad \Rightarrow \quad \sqrt{x} = \frac{1}{2} \quad \Rightarrow \quad x = \frac{1}{4} \]

5. **Finding the Corresponding y-coordinate**:
- Substitute \(x = \frac{1}{4}\) into \(y = 2\sqrt{x}\):
\[ y = 2\sqrt{\frac{1}{4}} = 2 \cdot \frac{1}{2} = 1 \]
- Thus, \(C\) is at \(\left(\frac{1}{4}, 1\right)\).

6. **Calculating the Maximum Area**:
- Substitute \(x = \frac{1}{4}\) back into the area formula:
\[ \text{Area} = 30 + 5\sqrt{\frac{1}{4}} - 5\left(\frac{1}{4}\right) \]
\[ = 30 + 5 \cdot \frac{1}{2} - \frac{5}{4} \]
\[ = 30 + \frac{5}{2} - \frac{5}{4} \]
\[ = 30 + \frac{10}{4} - \frac{5}{4} = 30 + \frac{5}{4} = 30 + 1.25 = 31.25 \]

### Final Answer:
The maximum area of triangle ACB is \(31.25\) square units.