To find the equation of the line on which the centroid of triangle PQR lies, where P, Q, and R are points on the parabola \( y^2 = 4ax \) and the normals at these points intersect at a single point, we can follow these steps:
### Step 1: Identify the points on the parabola
The points P, Q, and R on the parabola can be represented in terms of their parameters \( t_1, t_2, \) and \( t_3 \):
- Point P: \( P(t_1) = (a t_1^2, -2a t_1) \)
- Point Q: \( Q(t_2) = (a t_2^2, -2a t_2) \)
- Point R: \( R(t_3) = (a t_3^2, -2a t_3) \)
### Step 2: Write the equations of the normals
The slope of the tangent at a point on the parabola is given by \( \frac{1}{t} \), and the slope of the normal is \( -t \). Therefore, the equations of the normals at points P, Q, and R can be written as:
- Normal at P: \( y + 2a t_1 = -t_1 (x - a t_1^2) \)
- Normal at Q: \( y + 2a t_2 = -t_2 (x - a t_2^2) \)
- Normal at R: \( y + 2a t_3 = -t_3 (x - a t_3^2) \)
### Step 3: Rearranging the normal equations
Rearranging the equations of the normals gives us:
1. \( y = -t_1 x + (a t_1^3 + 2a t_1) \)
2. \( y = -t_2 x + (a t_2^3 + 2a t_2) \)
3. \( y = -t_3 x + (a t_3^3 + 2a t_3) \)
### Step 4: Find the intersection point of the normals
To find the intersection point of the three normals, we can set up a system of equations. However, we can also use the property that the sum of the slopes of the normals will give us the x-coordinate of the centroid.
### Step 5: Calculate the centroid of triangle PQR
The centroid \( G \) of triangle PQR is given by:
\[
G_x = \frac{x_P + x_Q + x_R}{3} = \frac{a(t_1^2 + t_2^2 + t_3^2)}{3}
\]
\[
G_y = \frac{y_P + y_Q + y_R}{3} = \frac{-2a(t_1 + t_2 + t_3)}{3}
\]
### Step 6: Determine the equation of the line on which the centroid lies
Since the y-coordinate of the centroid \( G_y \) is dependent on \( t_1 + t_2 + t_3 \), and the x-coordinate \( G_x \) is dependent on \( t_1^2 + t_2^2 + t_3^2 \), we can express the line on which the centroid lies as:
\[
y = 0 \quad \text{(since the centroid lies on the x-axis)}
\]
Thus, the equation of the line on which the centroid of triangle PQR lies is:
\[
3x + 2a = 0
\]
### Final Result
The equation of the line on which the centroid of triangle PQR lies is:
\[
3x + 2a = 0
\]
