Let AB be a chord of the parabola `x^(2) = 4y`. A circle drawn with AB as diameter passes through the vertex C of the parabola. If area of `DeltaABC = 20 sq`. units then the coordinates of A can be

Options

A. `(6,9)`
B. `(-6,9)`
C. `(-8,16)`
D. `(8,16)`

Explanation

To solve the problem step by step, we will follow the given information and derive the coordinates of point A on the parabola \( x^2 = 4y \).

### Step 1: Identify the Vertex and Points on the Parabola
The parabola given is \( x^2 = 4y \). The vertex \( C \) of this parabola is at the origin, \( (0, 0) \).

Let the points \( A \) and \( B \) on the parabola be represented in terms of parameters \( t_1 \) and \( t_2 \):
- Point \( A \) is \( (2t_1, t_1^2) \)
- Point \( B \) is \( (2t_2, t_2^2) \)

### Step 2: Area of Triangle ABC
The area \( \Delta \) of triangle \( ABC \) can be calculated using the determinant formula:
\[ \Delta = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates of points \( A \), \( B \), and \( C \):
\[ \Delta = \frac{1}{2} \left| 2t_1(t_2^2 - 0) + 2t_2(0 - t_1^2) + 0(t_1^2 - t_2^2) \right| \]
This simplifies to:
\[ \Delta = \frac{1}{2} \left| 2t_1 t_2^2 - 2t_2 t_1^2 \right| = \left| t_1 t_2 (t_2 - t_1) \right| \]

### Step 3: Set the Area Equal to 20
According to the problem, the area of triangle \( ABC \) is given as 20 square units:
\[ |t_1 t_2 (t_2 - t_1)| = 20 \]
This can be rewritten as:
\[ t_1 t_2 (t_2 - t_1) = \pm 20 \]

### Step 4: Solve for \( t_1 \) and \( t_2 \)
We can assume \( t_1 t_2 (t_2 - t_1) = 20 \) for simplicity. Rearranging gives us:
\[ t_1 t_2 (t_2 - t_1) = 20 \]

Let’s assume \( t_1 = 4 \) and \( t_2 = 1 \):
\[ 4 \cdot 1 \cdot (1 - 4) = 4 \cdot 1 \cdot (-3) = -12 \quad \text{(not valid)} \]
Now assume \( t_1 = 4 \) and \( t_2 = -1 \):
\[ 4 \cdot (-1) \cdot (-1 - 4) = 4 \cdot (-1) \cdot (-5) = 20 \quad \text{(valid)} \]
Thus, \( t_1 = 4 \) and \( t_2 = -1 \) is a valid pair.

### Step 5: Find Coordinates of A
Now substituting \( t_1 = 4 \) into the coordinates of point \( A \):
\[ A = (2t_1, t_1^2) = (2 \cdot 4, 4^2) = (8, 16) \]
Also substituting \( t_1 = -4 \) gives:
\[ A = (2 \cdot (-4), (-4)^2) = (-8, 16) \]

### Conclusion
The coordinates of point \( A \) can be \( (8, 16) \) or \( (-8, 16) \).