`y^(2)=4xandy^(2)=-8(x-a)` intersect at points A and C. Points O`(0,0)`, A,B `(a,0)`, and C are concyclic.
The area of cyclic quadrilateral OABC is

Options

A. (a) `24sqrt(3)`
B. (b) `48sqrt(2)`
C. (c) `12sqrt(6)`
D. (d) `18sqrt(5)`

Explanation

To solve the problem, we need to find the area of the cyclic quadrilateral OABC formed by the intersection points of the parabolas \(y^2 = 4x\) and \(y^2 = -8(x - a)\).

### Step-by-Step Solution:

1. **Identify the equations of the parabolas**:
- The first parabola is \(y^2 = 4x\).
- The second parabola is \(y^2 = -8(x - a)\).

2. **Set the equations equal to find points of intersection**:
\[ 4x = -8(x - a) \]
Rearranging gives:
\[ 4x + 8x - 8a = 0 \implies 12x = 8a \implies x = \frac{2a}{3} \]

3. **Substitute \(x\) back to find \(y\)**:
Using \(y^2 = 4x\):
\[ y^2 = 4 \left(\frac{2a}{3}\right) = \frac{8a}{3} \]
Thus, \(y = \pm \sqrt{\frac{8a}{3}}\).

4. **Determine the coordinates of points A and C**:
- Point A: \(\left(\frac{2a}{3}, \sqrt{\frac{8a}{3}}\right)\)
- Point C: \(\left(\frac{2a}{3}, -\sqrt{\frac{8a}{3}}\right)\)

5. **Coordinates of other points**:
- Point O: \((0, 0)\)
- Point B: \((a, 0)\)

6. **Calculate the lengths of diagonals AC and OB**:
- Length of \(AC\):
\[ AC = \sqrt{\left(\frac{2a}{3} - \frac{2a}{3}\right)^2 + \left(\sqrt{\frac{8a}{3}} - \left(-\sqrt{\frac{8a}{3}}\right)\right)^2} = \sqrt{0 + \left(2\sqrt{\frac{8a}{3}}\right)^2} = 2\sqrt{\frac{8a}{3}} \]

- Length of \(OB\):
\[ OB = \sqrt{(a - 0)^2 + (0 - 0)^2} = a \]

7. **Area of cyclic quadrilateral OABC**:
The area \(A\) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times AC \times OB = \frac{1}{2} \times 2\sqrt{\frac{8a}{3}} \times a = a\sqrt{\frac{8a}{3}} \]

8. **Find the value of \(a\)**:
Since points O, A, B, and C are concyclic, the angles subtended by the diameter must be \(90^\circ\). We can use the slopes of OA and AB to find \(a\).

- Slope of OA:
\[ m_{OA} = \frac{\sqrt{\frac{8a}{3}} - 0}{\frac{2a}{3} - 0} = \frac{\sqrt{\frac{8a}{3}}}{\frac{2a}{3}} = \frac{3\sqrt{\frac{8a}{3}}}{2a} = \frac{3\sqrt{8}}{2\sqrt{3}} = \frac{6\sqrt{2}}{2\sqrt{3}} = \sqrt{\frac{8}{3}} \cdot \frac{3}{2} \]

- Slope of AB:
\[ m_{AB} = \frac{\sqrt{\frac{8a}{3}} - 0}{\frac{2a}{3} - a} = \frac{\sqrt{\frac{8a}{3}}}{-\frac{a}{3}} = -3\sqrt{\frac{8a}{3a}} = -3\sqrt{\frac{8}{3}} \]

Setting the product of slopes equal to \(-1\):
\[ \sqrt{\frac{8}{3}} \cdot (-3\sqrt{\frac{8}{3}}) = -1 \]
Simplifying gives:
\[ -\frac{24}{3} = -1 \implies a = 12 \]

9. **Substituting \(a\) back to find the area**:
\[ \text{Area} = 12\sqrt{\frac{8 \times 12}{3}} = 12\sqrt{32} = 12 \cdot 4\sqrt{2} = 48\sqrt{2} \]

### Final Answer:
The area of the cyclic quadrilateral OABC is \(48\sqrt{2}\).