To solve the problem, we need to find the equation of the line on which the centroid of the triangle formed by the normals at points P, Q, and R of the parabola \( y^2 = 4ax \) lies. Let's break down the solution step by step.
### Step 1: Understand the Normal Equation
The equation of the normal to the parabola \( y^2 = 4ax \) at a point \( (at^2, 2at) \) is given by:
\[
y = mx - 2a(m - t)
\]
where \( m \) is the slope of the normal and \( t \) is the parameter corresponding to the point on the parabola.
### Step 2: Set Up Points P, Q, and R
Let the points P, Q, and R on the parabola be represented by parameters \( m_1, m_2, \) and \( m_3 \):
- Point P: \( (am_1^2, 2am_1) \)
- Point Q: \( (am_2^2, 2am_2) \)
- Point R: \( (am_3^2, 2am_3) \)
### Step 3: Find the Normals at Points P, Q, and R
Using the normal equation, we can write the normals at points P, Q, and R:
- Normal at P: \( y = m_1x - 2a(m_1 - m_1) \) simplifies to \( y = m_1x - 2a(0) \) or \( y = m_1x \)
- Normal at Q: \( y = m_2x - 2a(m_2 - m_2) \) simplifies to \( y = m_2x \)
- Normal at R: \( y = m_3x - 2a(m_3 - m_3) \) simplifies to \( y = m_3x \)
### Step 4: Find the Intersection of the Normals
Since the normals at points P, Q, and R meet at a point, we can express the intersection point in terms of the slopes \( m_1, m_2, \) and \( m_3 \). The intersection point can be found by solving the equations of the normals simultaneously.
### Step 5: Calculate the Centroid of Triangle PQR
The coordinates of the centroid \( G \) of triangle PQR can be calculated using the formula:
\[
G_x = \frac{x_1 + x_2 + x_3}{3}, \quad G_y = \frac{y_1 + y_2 + y_3}{3}
\]
Substituting the coordinates of points P, Q, and R:
\[
G_x = \frac{am_1^2 + am_2^2 + am_3^2}{3}, \quad G_y = \frac{2a(m_1 + m_2 + m_3)}{3}
\]
### Step 6: Use the Properties of Roots
Using Vieta's formulas for the roots \( m_1, m_2, m_3 \):
- The sum of the roots \( m_1 + m_2 + m_3 = 0 \)
- The product of the roots \( m_1m_2 + m_2m_3 + m_3m_1 = 2a - \frac{h}{a} \)
### Step 7: Substitute and Simplify
Substituting \( m_1 + m_2 + m_3 = 0 \) into the centroid's y-coordinate:
\[
G_y = \frac{2a(0)}{3} = 0
\]
Thus, the y-coordinate of the centroid is 0.
### Step 8: Equation of the Line
Since the y-coordinate of the centroid is 0, the equation of the line on which the centroid lies is:
\[
y = 0
\]
### Final Answer
The equation of the line on which the centroid of triangle PQR lies is:
\[
\boxed{y = 0}
\]
