椭圆的光学性质的推广

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A,B是任意点,C在x轴上,D在y轴上,E为CD上的动点.AP和BP分别等于E的横纵坐标.F在BP上,BF=OD.过A作AP垂线,过F作BP垂线交于G,求证GP恒与P的轨迹相切.

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设A(-1,0),B(1,0),则P满足方程$\frac{\sqrt{(x+1)^2+y^2}}{c}+\frac{\sqrt{(x-1)^2+y^2}}{d}=1$.去根号得$-4 c^2 d^2 x^2 y^2-2 c^4 d^2 x^2-2 c^2 d^2 x^4-2 c^2 d^4 x^2+4 c^2 d^2 x^2+4 c^4 d^2 x-4 c^2 d^4 x-2 c^4 d^2 y^2-2 c^2 d^2 y^4-2 c^2 d^4 y^2-4 c^2 d^2 y^2+c^4 d^4-2 c^4 d^2-2 c^2 d^4-2 c^2 d^2+2 c^4 x^2 y^2+c^4 x^4-4 c^4 x^3+6 c^4 x^2-4 c^4 x y^2-4 c^4 x+c^4 y^4+2 c^4 y^2+c^4+2 d^4 x^2 y^2+d^4 x^4+4 d^4 x^3+6 d^4 x^2+4 d^4 x y^2+4 d^4 x+d^4 y^4+2 d^4 y^2+d^4=0$
对其进行隐函数求导得P处的切线斜率$k=\frac{2 c^2 d^2 x^3+2 c^2 d^2 x y^2+c^4 d^2 x+c^2 d^4 x-2 c^2 d^2 x-c^4 d^2+c^2 d^4+c^4 \left(-x^3\right)+3 c^4 x^2-c^4 x y^2-3 c^4 x+c^4 y^2+c^4-d^4 x^3-3 d^4 x^2-d^4 x y^2-3 d^4 x-d^4 y^2-d^4}{y \left(-2 c^2 d^2 x^2-2 c^2 d^2 y^2+c^4 \left(-d^2\right)-c^2 d^4-2 c^2 d^2+c^4 x^2-2 c^4 x+c^4 y^2+c^4+d^4 x^2+2 d^4 x+d^4 y^2+d^4\right)}$

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令$k_a=\frac{y}{x+1},k_b=\frac{y}{x-1}$,只需证$\frac{\left(\frac{k-k_a}{k k_a+1}\right)^2+1}{\left(\frac{k-k_b}{k k_b+1}\right)^2+1}=\frac{d^2}{c^2}$
通分得$-d^2 \left(c^2 x^2-2 c^2 x+c^2 y^2+c^2-d^2 x^2-2 d^2 x-d^2 y^2-d^2\right) \left(-4 c^2 d^2 x^2 y^2-2 c^4 d^2 x^2-2 c^2 d^2 x^4-2 c^2 d^4 x^2+4 c^2 d^2 x^2+4 c^4 d^2 x-4 c^2 d^4 x-2 c^4 d^2 y^2-2 c^2 d^2 y^4-2 c^2 d^4 y^2-4 c^2 d^2 y^2+c^4 d^4-2 c^4 d^2-2 c^2 d^4-2 c^2 d^2+2 c^4 x^2 y^2+c^4 x^4-4 c^4 x^3+6 c^4 x^2-4 c^4 x y^2-4 c^4 x+c^4 y^4+2 c^4 y^2+c^4+2 d^4 x^2 y^2+d^4 x^4+4 d^4 x^3+6 d^4 x^2+4 d^4 x y^2+4 d^4 x+d^4 y^4+2 d^4 y^2+d^4\right)=0$
由P的轨迹方程知第二个因式为0,得证!