圆锥曲线中的存在性问题(2)

故(x1?x2)(x1?x2)?0. 于是由③式可得

(y1?y2)(y1?y2)??m2. ④

(x1?x2)(x1?x2)2yy?y2又Q,N,H三点共线,所以kQN?kQH,即1?1.

x1x1?x2y1y1?y21(y1?y2)(y1?y2)m2????于是由④式可得kPQ?kPH??.

x1x1?x22(x1?x2)(x1?x2)2而PQ?PH等价于kPQ?kPHm2??1,即???1,又m?0,得m?2, 22y2故存在m?2,使得在其对应的椭圆x??1上,对任意的k?0,都有

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