From a point (h,k) three normals are drawn to the parabola `y^2=4ax`. Tangents are drawn to the parabola at the of the normals to form a triangle. The Centroid of G of triangle is

Options

A. `((3a-h)/(3),0)`
B. `((3a-h)/(2),0)`
C. `((2a-h)/(3),0)`
D. `((a-h)/(2),0)`

Answer

To find the centroid \( G \) of the triangle formed by the tangents at the points where three normals drawn from a point \( (h, k) \) intersect the parabola \( y^2 = 4ax \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Parabola and Normals**:
The equation of the parabola is given by \( y^2 = 4ax \). The normals drawn from the point \( (h, k) \) will intersect the parabola at three points.

2. **General Points on the Parabola**:
The points on the parabola can be expressed in terms of parameters \( t_1, t_2, t_3 \):
- Point \( A \): \( (at_1^2, 2at_1) \)
- Point \( B \): \( (at_2^2, 2at_2) \)
- Point \( C \): \( (at_3^2, 2at_3) \)

3. **Conditions for Normals**:
The normals from the point \( (h, k) \) to the parabola satisfy three conditions:
- \( t_1 + t_2 + t_3 = 0 \) (Condition 1)
- \( t_1t_2 + t_2t_3 + t_3t_1 = \frac{2a - h}{a} \) (Condition 2)
- \( t_1t_2t_3 = \frac{k}{a} \) (Condition 3)

4. **Finding the Centroid**:
The centroid \( G \) of triangle \( ABC \) is given by:
\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]
Substituting the coordinates of points \( A, B, C \):
\[ G_x = \frac{at_1^2 + at_2^2 + at_3^2}{3} \]
\[ G_y = \frac{2at_1 + 2at_2 + 2at_3}{3} \]

5. **Simplifying the Coordinates**:
Using the conditions:
- For \( G_y \):
\[ G_y = \frac{2a(t_1 + t_2 + t_3)}{3} = \frac{2a(0)}{3} = 0 \]
- For \( G_x \):
\[ G_x = \frac{a(t_1^2 + t_2^2 + t_3^2)}{3} \]
We can use the identity \( t_1^2 + t_2^2 + t_3^2 = (t_1 + t_2 + t_3)^2 - 2(t_1t_2 + t_2t_3 + t_3t_1) \):
\[ t_1^2 + t_2^2 + t_3^2 = 0^2 - 2\left(\frac{2a - h}{a}\right) = -\frac{4a - 2h}{a} \]
Thus,
\[ G_x = \frac{a\left(-\frac{4a - 2h}{a}\right)}{3} = \frac{-(4a - 2h)}{3} = \frac{2h - 4a}{3} \]

6. **Final Coordinates of the Centroid**:
Therefore, the coordinates of the centroid \( G \) are:
\[ G = \left( \frac{2h - 4a}{3}, 0 \right) \]

### Conclusion:
The centroid \( G \) of the triangle formed by the tangents at the points where the normals intersect the parabola is given by:
\[ G = \left( \frac{2a - h}{3}, 0 \right) \]