This problem is a comprehensive test of your understanding of parabolas, straight lines, quadratic equations, and coordinate geometry, specifically the concept of a centroid. The key is to systematically break down the problem using standard formulas and algebraic manipulation.

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1. Problem Setup and Goal

We are given the following information:

• A parabola with equation: $y^2 = 20x$
• A line with equation: $y = mx+c$
• The line intersects the parabola at two distinct points, $P$ and $Q$.
• The point $R$ is the focus of the parabola.
• The centroid of triangle $PQR$ is $G(10,10)$.
• A relationship between the slope and y-intercept of the line: $c-m=6$.

Our ultimate goal is to determine the square of the distance between points $P$ and $Q$, denoted as $(PQ)^2$.

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2. Step-by-Step Solution

2.1. Determine the Focus of the Parabola, R

• Concept: The standard equation of a parabola opening to the right is $y^2 = 4ax$. The focus of such a parabola is located at the point $(a,0)$.
• Why this step? The coordinates of point $R$ are essential for applying the centroid formula for triangle $PQR$. We must first find these coordinates from the given parabola equation.
• Working: The given parabola equation is $y^2 = 20x$. We compare this with the standard form $y^2 = 4ax$. By equating the coefficients of $x$, we get: $4a = 20$ $a = 5$ Therefore, the focus $R$ of the parabola is $(a,0)$, which means $R = (5,0)$.

2.2. Formulate a Quadratic Equation for the Intersection Points P and Q

• Concept: To find the coordinates of the intersection points of a line and a parabola, we solve their equations simultaneously. This process typically leads to a quadratic equation whose roots represent the coordinates (either x or y) of these intersection points.
• Why this step? Let $P=(x_1, y_1)$ and $Q=(x_2, y_2)$. The sum and product of the roots of this quadratic equation will provide relationships between $x_1, x_2$ and $y_1, y_2$, which are crucial for using the centroid information and ultimately calculating $(PQ)^2$.
• Working: We have the line equation: $y = mx+c$ And the parabola equation: $y^2 = 20x$ It is often convenient to substitute one variable from the linear equation into the quadratic one. From the line equation, we can express $x$ in terms of $y$ (assuming $m \neq 0$; if $m=0$, the line is horizontal, $y=c$, leading to a single point of intersection with the parabola unless it's tangent or doesn't intersect, which contradicts the problem statement of two points $P$ and $Q$): $x = \frac{y-c}{m}$ Now, substitute this expression for $x$ into the parabola equation: $y^2 = 20 \left(\frac{y-c}{m}\right)$ Multiply both sides by $m$ to clear the denominator: $my^2 = 20(y-c)$ $my^2 = 20y - 20c$ Rearrange this into the standard quadratic form $Ay^2 + By + C = 0$: $my^2 - 20y + 20c = 0 \quad \ldots(1)$ Let $y_1$ and $y_2$ be the $y$-coordinates of points $P$