椭圆x^2|2+y^2=1,o是坐标原点,A,B,C是椭圆上三不同点,向量OA+OB+oc=0,证明AB与OC斜率之积为定值

Question

椭圆x^2|2+y^2=1,o是坐标原点,A,B,C是椭圆上三不同点,向量OA+OB+oc=0,证明AB与OC斜率之积为定值

Answer 1

设A(x1,y1),B(x2,y2),C(x3,y3)
则由OA+OB+OC=0可得
x1+x2+x3=0即x3=-(x1+x2)
y1+y2+y3=0即y3=-(y1+y2)
AB斜率=(y2-y1）/(x2-x1)
OC斜率=y3/x3=(y1+y2)/(x1+x2)
两者乘积=[(y2-y1)/(x2-x1)]*[(y1+y2)/(x1+x2)]=(y2²-y1²)/(x2²-x1²)
因为A、B在椭圆上,所以
x1²/2+y1²=1
x2²/2+y2²=1
两式相减得y2²-y1²+1/2*（x2²-x1²）=0
所以(y2²-y1²)/(x2²-x1²)=-1/2为定值.