直线y=kx+m(m≠0)与椭圆W:+y2=1相交于A,C两点,O是坐标原点. (1

直线ykxm(m0)与椭圆Wy21相交于AC两点,O是坐标原点.

(1)当点B的坐标为(0,1),且四边形OABC为菱形时,求AC的长;

(2)当点BW上且不是W的顶点时,证明:四边形OABC不可能为菱形.

答案

(1)因为四边形OABC为菱形,

所以ACOB相互垂直平分.

所以可设A(t),代入椭圆方程得1

t±.

所以|AC|2.

(2)假设四边形OABC为菱形.

因为点B不是W的顶点,且ACOB,所以k0.

y并整理得

(14k2)x28kmx4m240.

A(x1y1)C(x2y2),则

所以AC的中点为M

因为MACOB的交点,且m0k0

所以直线OB的斜率为-.

因为k·()1,所以ACOB不垂直.

所以OABC不是菱形,与假设矛盾.

所以当点B不是W的顶点时,四边形OABC不可能是菱形.

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