抛物线y=x2+px+q与x轴交于A,B两点,交y轴负半轴于点C,角ACB=90°,且OA分之1减去OB分之1等于OC分之2,求AB的长

1/OA-1/OB=2/OC
(OB-OA)/OA*OB=2/OC
OC=2OA*OB/(OB-OA)
角ACB=90° OA*OB=OC方=|Q|=OC Q=-1
OC=2OC方/(OB-OA) 2OC=OB-OA
(OB-OA)方=(X1-X2)方=(X1+X2)方-4X1X2=P方-4Q=4Q方=4
4Q方=P方-4Q P方=4Q方+4Q
AB的长=2