设AB是椭圆的x^2/a^2 + y^2/b^2=1的不垂直于对称轴且不过原点的弦,M为AB的中点,O为坐标原点,则

答：-b^2/a^2为定值,分析：用点差法.设A(X1,Y1),B(X2,Y2),AB中点M(XO,YO),则X1+X2=2XO,Y1+Y2=2YO,A,B在曲线上有X1^2/a^2+Y1^2/b^2=1,X2^2/a^2+Y2^2/b^2=1,当AB斜率K(AB)存在时两式相减有K(AB)=(Y1-Y2)/(X1-X2)=-(b^2/a^2)(X1+X2)/(Y1+Y2)=-(b^2/a^2)(Xo/YO),而OM斜率为K(OM)=(YO-0)/(XO-0)=YO/XO于是K(AB)*K(OM)=-(b^2/a^2)(Xo/YO)(YO/XO)=-b^2/a^2.解毕.