if P is a point on parabola `y^2=4ax` such that subtangents and subnormals at P are equal, then the coordinates of P are:

Options

A. (a, 2a) or (a, -2a)
B. `(2a, 2sqrt2a) or (2a, -2sqrt2a)`
C. `(4a, -4a) or (4a, 4a)`
D. none of these

Answer

To solve the problem, we need to find the coordinates of the point \( P \) on the parabola \( y^2 = 4ax \) such that the subtangents and subnormals at \( P \) are equal.

### Step-by-Step Solution:

1. **Identify the Point on the Parabola**:
The general point \( P \) on the parabola \( y^2 = 4ax \) can be represented as \( (at^2, 2at) \), where \( t \) is a parameter.

2. **Find the Derivative**:
To find the slopes needed for the subtangent and subnormal, we differentiate the equation of the parabola:
\[ y^2 = 4ax \implies 2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{2a}{y} \]
At the point \( P(at^2, 2at) \), we substitute \( y = 2at \):
\[ \frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t} \]

3. **Calculate the Subtangent**:
The length of the subtangent at point \( P \) is given by:
\[ \text{Subtangent} = y \cdot \frac{dx}{dy} = y \cdot \frac{1}{\frac{dy}{dx}} = y \cdot t = 2at \cdot t = 2at^2 \]

4. **Calculate the Subnormal**:
The length of the subnormal at point \( P \) is given by:
\[ \text{Subnormal} = y \cdot \frac{dy}{dx} = y \cdot \frac{1}{t} = 2at \cdot \frac{1}{t} = 2a \]

5. **Set Subtangent Equal to Subnormal**:
According to the problem, the subtangent and subnormal are equal:
\[ 2at^2 = 2a \]
Dividing both sides by \( 2a \) (assuming \( a \neq 0 \)):
\[ t^2 = 1 \]

6. **Solve for \( t \)**:
Taking the square root gives:
\[ t = 1 \quad \text{or} \quad t = -1 \]

7. **Find Coordinates of \( P \)**:
Substitute \( t = 1 \) and \( t = -1 \) back into the point \( P(at^2, 2at) \):
- For \( t = 1 \):
\[ P(1^2a, 2 \cdot 1 \cdot a) = (a, 2a) \]
- For \( t = -1 \):
\[ P((-1)^2a, 2 \cdot (-1) \cdot a) = (a, -2a) \]

### Final Coordinates of \( P \):
The coordinates of \( P \) are:
\[ P = (a, 2a) \quad \text{or} \quad P = (a, -2a) \]