Question

已知点P(1,1）为椭圆C :x^2/9+y^2/4=1内一定点,过点P的弦AB在点P被平分,求弦AB所在直线的方程.

Answer 1

设A（X1,y1).B(x2,y2)因为AB在椭圆上,有（x1)^2/9+(y1)^2/4=1 （x2)^2/9+(y2)^2/4=1 P(1.1)为AB中点,有（x1+x2)/2=1 (y1+y2)/2=1 即 x1+x2=2 y1+y2

即（x1)^2/9+(y1)^2/4=（x2)^2/9+(y2)^2/4 有（x1+x2)(x1-x2)/9=-(y1+y2）（y1-y2)/4 k(斜率）=-4/9(x1+x2)/(y1+y2) = -4/9 ×1/1 直线为y=-4/9(x-1)+1