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

To solve for $$PQ^2$$, we use the properties of the parabola and the centroid:

• Focus $$R$$: For $$y^2 = 20x$$, $$4a = 20 \implies a = 5$$. Thus, $$R = (5, 0)$$.
• Centroid $$G(10, 10)$$: Let $$P(x_1, y_1)$$ and $$Q(x_2, y_2)$$.
  • $$x_1 + x_2 + 5 = 3(10) \implies \mathbf{x_1 + x_2 = 25}$$
  • $$y_1 + y_2 + 0 = 3(10) \implies \mathbf{y_1 + y_2 = 30}$$

Substitute the points into $$y = mx + c$$:

$$(y_1 + y_2) = m(x_1 + x_2) + 2c \implies 30 = 25m + 2c$$

Given $$c - m = 6 \implies c = m + 6$$.

$$30 = 25m + 2(m + 6) \implies 18 = 27m \implies \mathbf{m = \frac{2}{3}, c = \frac{20}{3}}$$

The distance $$PQ$$ on a line with slope $$m$$ is $$PQ^2 = (x_1 - x_2)^2(1 + m^2)$$.

Find the intersection by substituting the line into $$y^2 = 20x$$:

$$\left[\frac{2}{3}(x + 10)\right]^2 = 20x \implies \frac{4}{9}(x^2 + 20x + 100) = 20x$$

$$x^2 - 25x + 100 = 0$$

Using roots $$(x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1x_2$$:

$$(x_1 - x_2)^2 = 25^2 - 4(100) = \mathbf{225}$$

Final Result:

$$PQ^2 = 225 \left(1 + \left(\frac{2}{3}\right)^2\right) = 225 \left(\frac{13}{9}\right) = 25 \times 13 = \mathbf{325}$$