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由雙曲線的方程可得a=3,b=1,c=10,∴F1(-10,0),F2(10,0).設點P(m,n),則 ;m29−n2=1①.設△PF1F2的重心G(x,y),則由三角形的重心座標公式可得x=m+10−103,y=n+0+03,即m=3x,n=3y,代入…
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