The vertex V of the parabola `y ^(2) = 4ax` and two points on the parabola together with a point P form a square. The coordinates of P are

Options

A. `(8a, 0)`
B. `(4a, 4a)`
C. `(8a,-4a)`
D. `(4a,0)`

Answer

To find the coordinates of point P such that the vertex V of the parabola \( y^2 = 4ax \) and two points on the parabola form a square, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Vertex and Points on the Parabola**:
The vertex \( V \) of the parabola \( y^2 = 4ax \) is at the origin, \( (0, 0) \). The points on the parabola can be represented parametrically as:
- Point \( Q_1(t_1) = (at_1^2, 2at_1) \)
- Point \( Q_2(t_2) = (at_2^2, 2at_2) \)

2. **Establish the Condition for a Square**:
For points \( V \), \( Q_1 \), \( Q_2 \), and \( P \) to form a square, the diagonals \( VQ_1 \) and \( Q_2P \) must be equal in length and perpendicular to each other.

3. **Calculate the Midpoint of \( V \) and \( Q_1 \)**:
The midpoint \( M_{VQ_1} \) of segment \( VQ_1 \) is:
\[ M_{VQ_1} = \left( \frac{0 + at_1^2}{2}, \frac{0 + 2at_1}{2} \right) = \left( \frac{at_1^2}{2}, at_1 \right) \]

4. **Calculate the Midpoint of \( Q_2 \) and \( P \)**:
Let the coordinates of point \( P \) be \( (x, y) \). The midpoint \( M_{Q_2P} \) is:
\[ M_{Q_2P} = \left( \frac{at_2^2 + x}{2}, \frac{2at_2 + y}{2} \right) \]

5. **Set the Midpoints Equal**:
Since the midpoints must be equal for the square configuration, we set:
\[ \frac{at_1^2}{2} = \frac{at_2^2 + x}{2} \quad \text{(1)} \]
\[ at_1 = \frac{2at_2 + y}{2} \quad \text{(2)} \]

6. **Solve for \( x \) and \( y \)**:
From equation (1):
\[ at_1^2 = at_2^2 + x \implies x = at_1^2 - at_2^2 = a(t_1^2 - t_2^2) \]
From equation (2):
\[ 2at_1 = 2at_2 + y \implies y = 2at_1 - 2at_2 = 2a(t_1 - t_2) \]

7. **Determine the Coordinates of Point \( P \)**:
Thus, the coordinates of point \( P \) are:
\[ P = \left( a(t_1^2 - t_2^2), 2a(t_1 - t_2) \right) \]

### Final Result:
The coordinates of point \( P \) are:
\[ P = \left( a(t_1^2 - t_2^2), 2a(t_1 - t_2) \right) \]