若椭圆x^2/9+y^2/4=1的弦AB被点P(1,1)平分,则AB所在直线的方程是_______.
问题描述:
若椭圆x^2/9+y^2/4=1的弦AB被点P(1,1)平分,则AB所在直线的方程是_______.
答
椭圆x^2/9+y^2/4=1的弦AB被点P(1,1)平分,则xA+xB=2xP=2,yA+yB=2yP=2k(AB)=(yA-yB)/(xA-xB)(xA)^2/9+(yA)^2/4=1.(1)(xB)^2/9+(yB)^2/4=1.(2)(1)-(2):(xA+xB)*(xA-xB)/9+(yA+yB)*(yA-yB)/4=0(xA+xB)/9+[(yA+yB)/4]*[(y...