Question

Let R be the focus of the parabola y² = 20x and the line y=mx+c intersect the parabola at two points Pand Q. Let the point G(10,10) be the centroid of the triangle PQR. If c-m=6, then (PQ)² is

• A. 296
• B. 325
• C. 317
• D. 346

Hint:

When solving geometry problems involving centroids, use symmetry and the parabola's focus-directrix property.

Answer 1

The Correct Option is B

We are given the parabola equation \(y^2 = 20x\) and the line equation \(y = mx + c\). We need to find the length of the chord \(PQ\) formed by the intersection of these two curves. 
Step 1: Find the intersection of the parabola and the line.
Substitute the equation of the line \(y = mx + c\) into the equation of the parabola \(y^2 = 20x\): \[ (mx + c)^2 = 20x. \] Expanding the equation: \[ m^2x^2 + 2mcx + c^2 = 20x. \] Rearranging the terms: \[ m^2x^2 + (2mc - 20)x + c^2 = 0. \] This is a quadratic equation in \(x\), and the points of intersection correspond to the roots of this equation. 
Step 2: Use the centroid information.
The centroid \(G(10, 10)\) is given as the centroid of the triangle \(PQR\). Using the centroid formula: \[ \left( \frac{x_1 + x_2 + 5}{3}, \frac{y_1 + y_2 + 0}{3} \right) = (10, 10), \] where \(R(5, 0)\) is the focus of the parabola and the points \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) are the points of intersection of the line and the parabola. This gives us two equations: 1. \(\frac{x_1 + x_2 + 5}{3} = 10 \quad \Rightarrow \quad x_1 + x_2 = 25\), 2. \(\frac{y_1 + y_2}{3} = 10 \quad \Rightarrow \quad y_1 + y_2 = 30\). 
Step 3: Solve for \(m\) and \(c\).
We are given that \(c - m = 6\), so \(c = m + 6\). To find the value of \(m\), use the relation derived from the intersection points. First, substitute \(y = mx + c\) into the equation of the parabola: \[ y^2 = 20\left( \frac{y - c}{m} \right). \] Rearrange and simplify: \[ y^2 - \frac{20y}{m} + \frac{20c}{m} = 0. \] Substitute \(c = m + 6\) into the equation: \[ \frac{20}{m} = 30 \quad \Rightarrow \quad m = \frac{2}{3}. \] Now substitute \(m = \frac{2}{3}\) into \(c = m + 6\) to find \(c\): \[ c = \frac{2}{3} + 6 = \frac{20}{3}. \] 
Step 4: Calculate the Points of Intersection.
Substitute \(m = \frac{2}{3}\) and \(c = \frac{20}{3}\) into the quadratic equation. The equation becomes: \[ y^2 - 30y + 200 = 0. \] Solving this quadratic equation, we get the roots: \[ y_1 = 10, \quad y_2 = 20. \] For \(y = 10\), substitute back into the line equation \(y = mx + c\) to get \(x = 5\). Thus, point \(P(5, 10)\) is one intersection point. For \(y = 20\), substitute back into the line equation \(y = mx + c\) to get \(x = 20\). Thus, point \(Q(20, 20)\) is the other intersection point. 
Step 5: Calculate the Length of the Chord \(PQ\).
The distance between the points \(P(5, 10)\) and \(Q(20, 20)\) is given by the distance formula: \[ PQ^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 = (20 - 5)^2 + (20 - 10)^2 = 15^2 + 10^2 = 225 + 100 = 325. \] 
Final Answer: 325.