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設直線方程為:y=2x+b 把直線帶入雙曲線x²;-y²;=12 得:3x²;+4bx+b²;+12=0 由於有解,判別式>0,即16b²;>12(b²;+12),b>6或者<-6 P1P2中點的橫坐標x=(x1+x2)/2=-2b/3,x>4或者x<-4 P1P2中點的縱坐標y=-b/3
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