From a variable point R on the line y = 2x + 3 tangents are drawn to the parabola `y^(2)=4ax` touch it at P and Q point. Find the locus of the centroid of the triangle PQR.

Answer

To find the locus of the centroid of triangle PQR formed by points P and Q on the parabola \(y^2 = 4ax\) and point R on the line \(y = 2x + 3\), we can follow these steps:

### Step 1: Identify the points P and Q
The points P and Q are the points of tangency from point R to the parabola. The coordinates of point P can be expressed as:
\[ P(t_1) = (at_1^2, 2at_1) \]
and for point Q:
\[ Q(t_2) = (at_2^2, 2at_2) \]

### Step 2: Find the coordinates of point R
Point R lies on the line \(y = 2x + 3\). We can express the coordinates of point R in terms of \(t_1\) and \(t_2\):
\[ R = (x_R, y_R) = (a t_1 t_2, a(t_1 + t_2)) \]

### Step 3: Calculate the centroid of triangle PQR
The centroid (G) of triangle PQR can be calculated using the formula:
\[ G = \left(\frac{x_P + x_Q + x_R}{3}, \frac{y_P + y_Q + y_R}{3}\right) \]
Substituting the coordinates of P, Q, and R:
\[ G = \left(\frac{at_1^2 + at_2^2 + at_1 t_2}{3}, \frac{2at_1 + 2at_2 + a(t_1 + t_2)}{3}\right) \]
This simplifies to:
\[ G = \left(\frac{a(t_1^2 + t_2^2 + t_1 t_2)}{3}, \frac{2a(t_1 + t_2) + a(t_1 + t_2)}{3}\right) \]
\[ G = \left(\frac{a(t_1^2 + t_2^2 + t_1 t_2)}{3}, \frac{3a(t_1 + t_2)}{3}\right) \]
\[ G = \left(\frac{a(t_1^2 + t_2^2 + t_1 t_2)}{3}, a(t_1 + t_2)\right) \]

### Step 4: Express \(t_1 + t_2\) and \(t_1 t_2\)
Using the properties of the tangents, we know:
- \(t_1 + t_2 = k/a\) (where k is the y-coordinate of the centroid)
- \(t_1 t_2 = \frac{(k - 3a)}{2a}\)

### Step 5: Substitute into the centroid coordinates
Substituting \(t_1 + t_2\) and \(t_1 t_2\) into the coordinates of G:
\[ G = \left(\frac{a\left(\left(\frac{k}{a}\right)^2 - 2\frac{(k - 3a)}{2a}\right)}{3}, k\right) \]
This simplifies to:
\[ G = \left(\frac{a\left(\frac{k^2}{a^2} - \frac{(k - 3a)}{a}\right)}{3}, k\right) \]

### Step 6: Find the locus
To find the locus, we eliminate the parameters \(t_1\) and \(t_2\). We can derive the equation by substituting \(x = h\) and \(y = k\) into the equation derived from the centroid coordinates:
\[ 6ah = k^2 - ak + 3a \]
Rearranging gives us:
\[ k^2 - ak - 6ah + 3a = 0 \]
This is a quadratic equation in \(k\).

### Final Equation
The locus of the centroid of triangle PQR can be expressed as:
\[ y^2 - 6ax - ay + 3a = 0 \]