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如圖所示,設點M(x,y),由條件可得,AB=4,EC=2,由點到直線的距離公式可得,MA2=(3x−y)210,MC2=(3x+y)210由垂徑定理可得,MA2+AB2=MC2+EC2,∴(3x−y)210+16=(3x+y)210+4,化簡可得,xy=10.∴點M的軌跡方程為xy=10.
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