解:设椭圆为x^2/a^2+y^2/b^2=1,直线AB的中点为M(已知),则
xA+xB=2xM,yA+yB=2yM
b^2*(xA^2-xB^2)+a^2*(yA^2-yB^2)=0
b^2*(xA+xB)*(xA-xB)+a^2*(yA+yB)*(yA-yB)=0
b^2*2xM*(xA-xB)+a^2*2yM*(yA-yB)=0
k=k(AB)=(yA-yB)/(xA-xB)=-b^2*xM/(a^2*yM)
直线AB:y-yM=k*(x-xM)与椭圆方程解得交点坐标A( xA,yA),B ( xB,yB)
AB^2=(xA-xB)^2+(yA-yB)^2
|AB|=