温新堂个性化一对一教学
C:x2?y2?R2(1 ?y?kx?t?由（2）知?x2得x2?4(kx?t)2?4, 2?y?1??4t1?k2, 即t2?R2(1?k2) ①, 即(1?4k2)x2?8ktx?4t2?4?0有唯一解 则△=64k2t2?16(1?4k2)(t2?1)?16(4k2?t2?1)?0, 即4k2?t2?1?0, ② 2?23R?t?2?4?R由①②得?, 此时A,B重合为B1(x1,y1)点, w.w.w.k.s.5.u.c.o.m 2?k2?R?12??4?R8kt?x?x??12222?4t?416R?16?1?4k2由? 中x1?x2,所以,x1?, ?2221?4k3R?xx?4t?4122?1?4k?2B1(x1,y1)点在椭圆上,所以y1?1?142x1?24?R3R222,所以|OB1|?x1?y1?5?22224R2, 2在直角三角形OA1B1中,|A1B1|?|OB1|?|OA1|?5?4R24R22?R?5?(24R2?R)因为 ?R?4当且仅当R?22?(1,2)时取等号,所以|A1B1|?5?4?1,即 当R? 2?(1,2)时|A1B1|取得最大值,最大值为1. 16 一切为了孩子--------温新堂教育