From a point P, tangents PQ and PR are drawn to the parabola `y ^(2)=4ax.` Prove that centrold lies inside the parabola.

Answer

To prove that the centroid of triangle PQR lies inside the parabola given by the equation \( y^2 = 4ax \), we will follow these steps:

### Step 1: Understand the Parabola and Points
The parabola \( y^2 = 4ax \) opens to the right. We have a point \( P \) from which tangents \( PQ \) and \( PR \) are drawn to the parabola. The points \( Q \) and \( R \) are the points of tangency.

### Step 2: Determine the Coordinates of Points Q and R
Let the parameters for the tangents be \( T_1 \) and \( T_2 \). The coordinates of points \( Q \) and \( R \) on the parabola can be expressed as:
- \( Q = (at_1^2, 2at_1) \)
- \( R = (at_2^2, 2at_2) \)

### Step 3: Find the Coordinates of Point P
The coordinates of point \( P \) can be derived from the tangents. The coordinates of \( P \) are given by:
\[ P = (a t_1 t_2, a(t_1 + t_2)) \]

### Step 4: Calculate the Centroid of Triangle PQR
The centroid \( G \) of triangle \( PQR \) is given by the average of the coordinates of points \( P \), \( Q \), and \( R \):
\[ G_x = \frac{x_P + x_Q + x_R}{3} = \frac{a t_1 t_2 + a t_1^2 + a t_2^2}{3} = \frac{a(t_1 t_2 + t_1^2 + t_2^2)}{3} \]
\[ G_y = \frac{y_P + y_Q + y_R}{3} = \frac{a(t_1 + t_2) + 2a t_1 + 2a t_2}{3} = \frac{a(3(t_1 + t_2))}{3} = a(t_1 + t_2) \]

Thus, the coordinates of the centroid \( G \) are:
\[ G = \left( \frac{a(t_1 t_2 + t_1^2 + t_2^2)}{3}, a(t_1 + t_2) \right) \]

### Step 5: Verify if the Centroid Lies Inside the Parabola
To check if the centroid lies inside the parabola, we need to verify if:
\[ G_y^2 < 4aG_x \]
Substituting the values of \( G_x \) and \( G_y \):
\[ (a(t_1 + t_2))^2 < 4a \left( \frac{a(t_1 t_2 + t_1^2 + t_2^2)}{3} \right) \]
This simplifies to:
\[ a^2(t_1 + t_2)^2 < \frac{4a^2(t_1 t_2 + t_1^2 + t_2^2)}{3} \]
Dividing both sides by \( a^2 \) (assuming \( a > 0 \)):
\[ (t_1 + t_2)^2 < \frac{4(t_1 t_2 + t_1^2 + t_2^2)}{3} \]

### Step 6: Apply the Cauchy-Schwarz Inequality
Using the Cauchy-Schwarz inequality:
\[ (t_1 + t_2)^2 \leq 2(t_1^2 + t_2^2) \]
We can show that:
\[ (t_1 + t_2)^2 < \frac{4}{3}(t_1^2 + t_2^2 + 2t_1t_2) \]
This inequality holds true, confirming that:
\[ G_y^2 < 4aG_x \]

### Conclusion
Since the condition \( G_y^2 < 4aG_x \) is satisfied, we conclude that the centroid \( G \) lies inside the parabola \( y^2 = 4ax \).