Let R be the focus of the parabola y^2=20x an | vedclass

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(A) The parabola is $y^2=20x$,so the focus $R$ is $(5, 0)$.Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ be the intersection points of $y=mx+c$ and $y^2=20x$.Su

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$325$

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(A) The parabola is $y^2=20x$,so the focus $R$ is $(5, 0)$.Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ be the intersection points of $y=mx+c$ and $y^2=20x$.Substituting $x = (y-c)/m$ into the parabola equation: $y^2 = 20(y-c)/m \Rightarrow my^2 - 20y + 20c = 0$.From the quadratic equation,$y_1+y_2 = 20/m$.The centroid $G(10, 10)$ of $	riangle PQR$ satisfies $(y_1+y_2+y_R)/3 = 10$. Since $y_R = 0$,we have $(y_1+y_2)/3 = 10 \Rightarrow y_1+y_2 = 30$.Thus,$20/m = 30 \Rightarrow m = 2/3$.Given $c-m=6$,we have $c = 6 + 2/3 = 20/3$.The quadratic equation for $y$ becomes $y^2 - 30y + 200 = 0 \Rightarrow (y-10)(y-20) = 0$.So $y_1=10$ and $y_2=20$. Corresponding $x$ values are $x_1 = y_1^2/20 = 100/20 = 5$ and $x_2 = y_2^2/20 = 400/20 = 20$.Thus $P(5, 10)$ and $Q(20, 20)$.$(PQ)^2 = (20-5)^2 + (20-10)^2 = 15^2 + 10^2 = 225 + 100 = 325$.

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 point $G(10, 10)$ be the centroid of the triangle $PQR$. If $c-m=6$,then $(PQ)^2$ is

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