(经典)高考数学直线和圆锥曲线常考题型及方法

则△=16k2m2?4(1?2k2)(2m2?8)?8(8k2?m2?4)?0,即8k2?m2?4?0

4km?x?x??12??1?2k2?22m?8?xx?12?1?2k2?k2(2m2?8)4k2m2m2?8k22y1y2?(kx1?m)(kx2?m)?kx1x2?km(x1?x2)?m???m?1?2k21?2k21?2k222,

要使

O?A,需使x1x2?2m2?8m2?8k222??03m?8k?8?0,所以,所以y1?y20,即221?2k1?2k?m2?283m2?8262622k??0又8k?m?4?0,所以?2,所以m2?,即m?或m??,因为直

3833?3m?82线y?kx?m为圆心在原点的圆的一条切线,所以圆的半径为

8m2m282622x?y?r???,,,所求的圆为,此时圆的切线r?r?2223m?831?k331?k1?82my?kx?m都满足m?262626或m??,而当切线的斜率不存在时切线为x??与椭圆333x2y226262626??1的两个交点为(,?)或(?,?)满足OA?OB,综上, 存在圆心在原843333点的圆x2?y2?，使得该圆的任意一条切线与椭圆E恒有两个交点A,B,且OA?OB.

4km?x?x??12??1?2k2因为?, 2?xx?2m?8?121?2k2?834km22m2?88(8k2?m2?4)所以(x1?x2)?(x1?x2)?4x1x2?(?, )?4??22221?2k1?2k(1?2k)22|AB|?(x1?x2)??y1?y2?228(8k2?m2?4) ?(1?k)(x1?x2)?(1?k)(1?2k2)2222324k4?5k2?132k2???[1?4], 34k4?4k2?134k?4k2?1①当k?0时|AB|?321[1?]

134k2?2?4k 17

因为4k2?所以

1?4?8所以0?2k11?, 14k2?2?48k32321?[1?]?12,

1334k2?2?4k426?|AB|?23当且仅当k??时取”=”. w.w.w.k.s.5.u.c.o.m 32所以② 当k?0时,|AB|?46. 32626262646, ,?)或(?,?),所以此时|AB|?33333③ 当AB的斜率不存在时, 两个交点为(综上, |AB |的取值范围为446?|AB|?23即: |AB|?[6,23] 33 18