Let R br the focus of the parabola `y^(2)=20x ` and the lines `y=mx+c` intersect the parabola at two points `P`and `Q` . let the point `G(10,10)` be the centroid of the triangle `PQR`. If `c-m=6then (PQ)^(2) ` is

Options

A. `296`
B. `317`
C. `325`
D. `346`

Answer

To solve the problem step by step, we will follow the given information and derive the necessary equations.

### Step 1: Identify the focus of the parabola
The equation of the parabola is given as \( y^2 = 20x \). The standard form of a parabola that opens to the right is \( y^2 = 4px \). Here, \( 4p = 20 \), which gives \( p = 5 \). Therefore, the focus \( R \) of the parabola is at the point \( (5, 0) \).

**Hint:** Remember that the focus of a parabola defined by \( y^2 = 4px \) is located at \( (p, 0) \).

### Step 2: Find the intersection points of the line and the parabola
The line is given by \( y = mx + c \). To find the intersection points \( P \) and \( Q \), substitute \( y \) into the parabola's equation:

\[ (mx + c)^2 = 20x \]

Expanding this gives:

\[ m^2x^2 + 2mcx + c^2 = 20x \]

Rearranging terms results in a quadratic equation:

\[ m^2x^2 + (2mc - 20)x + c^2 = 0 \]

**Hint:** Use the quadratic formula to find the roots \( x_1 \) and \( x_2 \) of this equation.

### Step 3: Use the centroid condition
The centroid \( G \) of triangle \( PQR \) is given as \( G(10, 10) \). The coordinates of the centroid are given by:

\[ G\left(\frac{x_1 + x_2 + 5}{3}, \frac{y_1 + y_2 + 0}{3}\right) = (10, 10) \]

From this, we can set up the equations:

1. \( \frac{x_1 + x_2 + 5}{3} = 10 \)
2. \( \frac{y_1 + y_2}{3} = 10 \)

From the first equation, we find:

\[ x_1 + x_2 + 5 = 30 \implies x_1 + x_2 = 25 \]

From the second equation, we find:

\[ y_1 + y_2 = 30 \]

**Hint:** The sum of the x-coordinates and y-coordinates of the points can help you find the individual coordinates.

### Step 4: Relate \( c \) and \( m \)
We are given that \( c - m = 6 \). Therefore, we can express \( c \) in terms of \( m \):

\[ c = m + 6 \]

**Hint:** Substitute this expression into the quadratic equation derived from the intersection points.

### Step 5: Substitute \( c \) into the quadratic equation
Substituting \( c = m + 6 \) into the quadratic equation gives:

\[ m^2x^2 + (2m(m + 6) - 20)x + (m + 6)^2 = 0 \]

This simplifies to:

\[ m^2x^2 + (2m^2 + 12m - 20)x + (m^2 + 12m + 36) = 0 \]

**Hint:** Collect like terms carefully when substituting.

### Step 6: Find \( PQ^2 \)
Using the formula for the distance between the points \( P \) and \( Q \):

\[ PQ^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2 \]

From the quadratic equation, we can find \( x_1 - x_2 \) using the discriminant. The discriminant \( D \) of the quadratic is given by:

\[ D = (2mc - 20)^2 - 4m^2c^2 \]

Using the relationships we derived, we can find \( PQ^2 \).

**Hint:** Remember that \( PQ^2 \) can also be expressed in terms of the coordinates of the centroid and the focus.

### Final Calculation
After substituting and simplifying, we find that:

\[ PQ^2 = 325 \]

Thus, the final answer is:

\[ \boxed{325} \]