如何证明椭圆上一点处切线为其与两焦点外角平分线

Question

如何证明椭圆上一点处切线为其与两焦点外角平分线

Answer 1

证明：不失一般性,设椭圆方程为x^2/a^2+y^2/b^2=1 (a>b>0),交点分别为F1(-c,0)、F2(c,0).不失一般性,设不与F1F2共线的椭圆第一象限上任意一点P(x0,y0),则有
c^2=a^2-b^2①
x0^2/a^2+y0^2/b^2=1②
由②得b^2x0^2+a^2y0^2=a^2b^2
x^2/a^2+y^2/b^2=1两边对x求导,得
2x/a^2+2yy'/b^2=0得y'=-b^2*x/(a^2*y)
则过点P的切线方程为y-y0=-b^2*x0/(a^2*y0)*(x-x0)
令y=0,解得x=(a^2y0^2+b^2x0^2)/(b^2x0)=a^2b^2/(b^2x0)=a^2/x0
则过椭圆上点P(x0,y0)的切线交x轴于点M(a^2/x0,0).于是
|F1M|=a^2/x0+c,|F2M|=a^2/x0-c
|PF1|=√[(x0+c)^2+y0^2],|PF2|=√[(x0-c)^2+y0^2]
则
(|F1M|*|PF2|)^2=(a^2/x0+c)^2*[(x0-c)^2+y0^2]
(|F2M|*|PF1|)^2=(a^2/x0-c)^2*[(x0+c)^2+y0^2]
于是：
(|F1M|*|PF2|)^2-(|F2M|*|PF1|)^2=(a^2/x0+c)^2*[(x0-c)^2+y0^2]-(a^2/x0-c)^2*[(x0+c)^2+y0^2]
=(a^4/x0^2+2a^2c/x0+c^2)(x0^2-2cx0+c^2+y0^2)-(a^4/x0^2-2a^2c/x0+c^2)(x0^2+2cx0+c^2+y0^2)
=2a^2c/x0*(x0^2-2cx0+c^2+y0^2)-2cx0*(a^4/x0^2+2a^2c/x0+c^2)-(-2a^2c/x0)*(x0^2+2cx0+c^2+y0^2)-2cx0*(a^4/x0^2-2a^2c/x0+c^2)
=2a^2c/x0*2(x0^2+c^2+y0^2)-2cx0*2(a^4/x0^2+c^2)
=4c*[a^2/x0*(x0^2+c^2+y0^2)-x0*(a^4/x0^2+c^2)]
=4c*[(a^2x0-c^2x0)+a^2/x0*(c^2-a^2)+a^2y0^2/x0]
=4c*[b^2x0-a^2b^2/x0+a^2y0^2/x0]=4c/x0*[b^2x0^2-a^2b^2+a^2y0^2]=0
故(|F1M|*|PF2|)^2=(|F2M|*|PF1|)^2
|F1M|*|PF2|=|F2M|*|PF1|
|PF1|/|PF2|=|F1M|/|F2M|
依外角平分线性质定理,知该切线平分焦点三角形PF1F2的外角.