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, where R is a variable point on the line \( y = 2x + 3 \) and tangents are drawn from R to the parabola \( y^2 = 4ax \), we can follow these steps:

### Step 1: Identify the coordinates of point R
Let the coordinates of point R be \( (x, y) \). Since R lies on the line \( y = 2x + 3 \), we can express y in terms of x:
\[ y = 2x + 3 \]

### Step 2: Write the equation of the tangents from point R to the parabola
The equation of the parabola is \( y^2 = 4ax \). The equation of the tangent to the parabola at point P, which has parameter \( t_1 \), is given by:
\[ yy_1 = 2a(x + x_1) \]
where \( (x_1, y_1) \) is the point of tangency. For point P, we have:
\[ x_1 = at_1^2, \quad y_1 = 2at_1 \]
Thus, the tangent at P becomes:
\[ y(2at_1) = 2a(x + at_1^2) \]
Simplifying this gives:
\[ yt_1 = x + at_1^2 \]

### Step 3: Find the coordinates of points P and Q
Similarly, for point Q with parameter \( t_2 \), we have:
\[ y(2at_2) = 2a(x + at_2^2) \]
This gives us the second tangent equation:
\[ yt_2 = x + at_2^2 \]

### Step 4: Find the intersection of the tangents
To find the coordinates of point R, we need to solve the two tangent equations simultaneously. We can express the intersection point of the tangents as:
\[ x = at_1t_2, \quad y = a(t_1 + t_2) \]

### Step 5: Find the centroid of triangle PQR
The coordinates of the centroid C of triangle PQR can be calculated using the formula:
\[ C\left( \frac{x_P + x_Q + x_R}{3}, \frac{y_P + y_Q + y_R}{3} \right) \]
Substituting the coordinates of P, Q, and R:
\[ C\left( \frac{at_1^2 + at_2^2 + at_1t_2}{3}, \frac{2at_1 + 2at_2 + a(t_1 + t_2)}{3} \right) \]
This simplifies to:
\[ C\left( \frac{a(t_1^2 + t_2^2 + t_1t_2)}{3}, \frac{2a(t_1 + t_2) + a(t_1 + t_2)}{3} \right) \]
\[ C\left( \frac{a(t_1^2 + t_2^2 + t_1t_2)}{3}, \frac{3a(t_1 + t_2)}{3} \right) \]
\[ C\left( \frac{a(t_1^2 + t_2^2 + t_1t_2)}{3}, a(t_1 + t_2) \right) \]

### Step 6: Express the parameters in terms of x and y
Since \( t_1 + t_2 = \frac{y - 3}{2a} \) and \( t_1t_2 = \frac{x}{a} \), we can substitute these into the centroid coordinates:
\[ h = \frac{a\left(\frac{y^2 - 6y + 9}{4a^2} + \frac{x}{a}\right)}{3}, \quad k = a\left(\frac{y - 3}{2a}\right) \]

### Step 7: Find the locus of the centroid
By eliminating \( t_1 \) and \( t_2 \) and expressing the relationship between h and k, we can derive the locus of the centroid. After simplification, we find:
\[ 6ax = 2y^2 - ay - 3a \]
This is the equation of a parabola.

### Final Result
The locus of the centroid of triangle PQR is given by:
\[ 6ax = 2y^2 - ay - 3a \]