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

Option: 1

325


Option: 2

346


Option: 3

296


Option: 4

317


Answers (1)

best_answer

\begin{aligned} & \mathrm{y}^2=20 x, y=m x+c \\ & \mathrm{y}^2=20\left(\frac{\mathrm{y}-\mathrm{c}}{\mathrm{m}}\right) \\ & \mathrm{y}^2-\frac{20 \mathrm{y}}{\mathrm{m}}+\frac{20 \mathrm{c}}{\mathrm{m}}=0 \quad \frac{\mathrm{y}_1+\mathrm{y}_2+\mathrm{y}_3}{3}=10 \\ & \frac{20}{\mathrm{~m}}=30 \\ & \mathrm{~m}=2 / 3 \\ & \text { and } \mathrm{c}-\mathrm{m}=6 \end{aligned}

\begin{aligned} & \mathrm{c}=\frac{2}{3}+6 \Rightarrow \frac{20}{3}=c \\ & y^2-30 \mathrm{y}+\frac{20 \times 20 / 3}{2 / 3}=0 \Rightarrow \quad \mathrm{y}^2-30 \mathrm{y}+200=0 \\ & \mathrm{y}=10, \mathrm{y}=20 \\ & \mathrm{y}=20, \mathrm{x}=20 \quad P(5,10) ;(20,20) \mathrm{Q} \\ & \frac{20+5+x}{3}=10 \Rightarrow x=5 \quad \mathrm{PQ}^2=15^2+10^2=225+100=325 \end{aligned}

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