如图平面上有A(1,0),B(-1,0)两点,已知圆的方程为(x-3)^2+(y-4)^2=4

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  • 题目中的圆的参数方程为:x = 3 + 2*cosθ ,y = 4 + 2*sinθ

    即圆上点P的坐标为(3 + 2*cosθ ,4 + 2*sinθ)

    因此|AP|2+|BP|2 = (2 + 2*cosθ)² + (4 + 2*sinθ)² + (4 + 2*cosθ)² + (4 + 2*sinθ)² = 60 + (8/5)*((3/5)*cosθ + (4/5)*sinθ)

    = 60 + (8/5)*sin(θ + a)

    其中tan(a) = 3/4 ,因此若要取得最小值,需要使得sin(θ + a)最小,即sin(θ + a) = -1,即

    (3/5)*cosθ + (4/5)*sinθ = -1 ,显然当cosθ = -(3/5) ,sinθ = -(4/5)的时候,达到最小值(或者将上式带入圆的方程中解得cosθ和sinθ)因此可得:

    x = 3 + 2*cosθ = 3 - 1.2 = 1.8

    y = 4 + 2*sinθ = 4 - 1.6 = 2.4