Tangent and normal at any point P of the parabola `y^(2)=4ax(a gt 0)` meet the x-axis at T and N respectively. If the lengths of sub-tangent and sub-normal at this point are equal, then the area of `DeltaPTN` is given by

Options

A. `4a^(2)`
B. `6sqrt2a^(2)`
C. `4sqrt2a^(2)`
D. none of these

Explanation

To solve the problem, we need to find the area of triangle PTN formed by the tangent and normal to the parabola \( y^2 = 4ax \) at a point \( P \) on the parabola. We are given that the lengths of the sub-tangent and sub-normal at point \( P \) are equal.

### Step-by-Step Solution:

1. **Identify the point P on the parabola:**
Let \( P(t) \) be a point on the parabola \( y^2 = 4ax \). The coordinates of point \( P \) can be represented as:
\[ P = (at^2, 2at) \]

2. **Find the equations of the tangent and normal at point P:**
The equation of the tangent at point \( P \) is given by:
\[ y = tx - at^2 + 2at \]
The equation of the normal at point \( P \) is:
\[ y = -\frac{1}{t}(x - at^2) + 2at \]

3. **Determine the points T and N where the tangent and normal meet the x-axis:**
To find point \( T \) (where the tangent meets the x-axis), set \( y = 0 \) in the tangent equation:
\[ 0 = tx - at^2 + 2at \implies tx = at^2 - 2at \implies x = \frac{at(t - 2)}{t} = a(t - 2) \]
Thus, \( T = (a(t - 2), 0) \).

For point \( N \) (where the normal meets the x-axis), set \( y = 0 \) in the normal equation:
\[ 0 = -\frac{1}{t}(x - at^2) + 2at \implies \frac{1}{t}(x - at^2) = 2at \implies x - at^2 = 2at^2 \implies x = 3at^2 \]
Thus, \( N = (3at^2, 0) \).

4. **Calculate the lengths of the sub-tangent and sub-normal:**
The length of the sub-tangent \( TG \) is given by:
\[ TG = \frac{y}{\frac{dy}{dx}} = \frac{2at}{t} = 2a \]
The length of the sub-normal \( GN \) is given by:
\[ GN = \frac{y}{\frac{dx}{dy}} = \frac{2at}{\frac{1}{t}} = 2at^2 \]

5. **Set the lengths equal:**
Given that the lengths of the sub-tangent and sub-normal are equal:
\[ 2a = 2at^2 \implies 1 = t^2 \implies t = 1 \quad (\text{since } a > 0) \]

6. **Substitute \( t = 1 \) back into the coordinates:**
The coordinates of point \( P \) become:
\[ P = (a, 2a) \]
The coordinates of points \( T \) and \( N \) become:
\[ T = (a(1 - 2), 0) = (-a, 0) \]
\[ N = (3a, 0) \]

7. **Calculate the area of triangle PTN:**
The area \( A \) of triangle \( PTN \) can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base \( TN \) is the distance between points \( T \) and \( N \):
\[ TN = 3a - (-a) = 4a \]
The height is the y-coordinate of point \( P \), which is \( 2a \):
\[ A = \frac{1}{2} \times 4a \times 2a = 4a^2 \]

### Final Answer:
The area of triangle \( PTN \) is \( 4a^2 \).