题目

△ABC中,B(-1,0)、C(1,0),并且∠C=2∠B,求顶点A的轨迹方程. 答案：解：设A(x,y),当A点在x轴上方时,tanB=,tan(π-2∠B)=,∴=-=-.化简得3x2-y2+2x-1=0(y＞0).当A点在x轴下方时,同理易得3x2-y2+2x-1=0(y＜0).又|AB|＞|AC|,∴x＞0.∴A点轨迹方程为3x2-y2+2x-1=0(x＞0,y≠0).

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