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=2
k(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]*[(yA-yB)/(xA-xB)]=0
2/9+(2/4)*k(AB)=0
k(AB)=-4/9
AB:y-1=(-4/9)*(x-1)
AB所在直线的方程是:4x+9y-13=0