如图,在平面直角坐标系xOy,顶点为M的抛物线y=ax²+bx(a>0)经过点A和x轴正半轴的点B,AO=BO

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  • AO=2 , ∠AOB=120ºA:(﹣1, √3) B:(2,0)y=ax²+bxa-b=√34a+2b=0 y= √3/3 x^2- 2√3/3x (2)y= √3/3 x^2- 2√3/3x顶点M:(1,﹣ √3/3)∴∠BOM=30°,∴∠AOM=30°+120°=150° (3)∠AOB=120º, ∠OBA=∠OAB=30ºOA≠OM∴∠OAM≠30º , ∠OMA≠30º∴C不在B的左侧

    ∠ABC=∠AOM=150º △ABC1∽△AOM∴AB/AO=BC1/MO AB=2√3, AO=2, MO=2√3/3∴BC1=2 ∴OC1=4,∴C1:(4,0) △C2BA∽△AOM ∴AB/MO=C2B/AO BC2=6,∴OC2=8,∴C2:(8,0) 点C的坐标为:(4,0)或(8,0).