To find the centroid \( G \) of the triangle formed by the tangents drawn from points \( A \), \( B \), and \( C \) on the parabola \( y^2 = 4ax \) from a point \( (h, k) \), we will follow these steps:
### Step 1: Parametric Representation of Points on the Parabola
The points \( A \), \( B \), and \( C \) on the parabola can be represented in parametric form as:
- \( A(t_1) = (at_1^2, 2at_1) \)
- \( B(t_2) = (at_2^2, 2at_2) \)
- \( C(t_3) = (at_3^2, 2at_3) \)
### Step 2: Finding the Coordinates of Points \( P \), \( Q \), and \( R \)
The points \( P \), \( Q \), and \( R \) are the intersection points of the tangents drawn at points \( A \), \( B \), and \( C \).
- The coordinates of point \( P \) (intersection of tangents at \( A \) and \( B \)):
\[
P = (a t_1 t_2, 2a(t_1 + t_2))
\]
- The coordinates of point \( Q \) (intersection of tangents at \( A \) and \( C \)):
\[
Q = (a t_1 t_3, 2a(t_1 + t_3))
\]
- The coordinates of point \( R \) (intersection of tangents at \( B \) and \( C \)):
\[
R = (a t_2 t_3, 2a(t_2 + t_3))
\]
### Step 3: Finding the Centroid \( G \)
The centroid \( G \) of triangle \( PQR \) is given by the formula:
\[
G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)
\]
Substituting the coordinates of points \( P \), \( Q \), and \( R \):
- \( x \)-coordinate of \( G \):
\[
G_x = \frac{a t_1 t_2 + a t_1 t_3 + a t_2 t_3}{3} = \frac{a(t_1 t_2 + t_1 t_3 + t_2 t_3)}{3}
\]
- \( y \)-coordinate of \( G \):
\[
G_y = \frac{2a(t_1 + t_2 + t_1 + t_3 + t_2 + t_3)}{3} = \frac{2a(2(t_1 + t_2 + t_3))}{3} = \frac{4a(t_1 + t_2 + t_3)}{3}
\]
### Step 4: Using Conditions for Co-normal Points
From the properties of co-normal points:
1. \( t_1 + t_2 + t_3 = 0 \)
2. \( t_1 t_2 + t_2 t_3 + t_3 t_1 = \frac{2a - h}{a} \)
### Step 5: Substitute Values in Centroid Coordinates
Using \( t_1 + t_2 + t_3 = 0 \):
- The \( y \)-coordinate becomes:
\[
G_y = \frac{4a(0)}{3} = 0
\]
- The \( x \)-coordinate becomes:
\[
G_x = \frac{a \left( \frac{2a - h}{a} \right)}{3} = \frac{2a - h}{3}
\]
### Final Result
Thus, the coordinates of the centroid \( G \) of triangle \( PQR \) are:
\[
G = \left( \frac{2a - h}{3}, 0 \right)
\]
