?3?又由于?x??(3y)2?9, ?2?2x2化简得M的轨迹C的方程为?y2?1.
4(2)设直线PQ的方程为y?kx?与C的方程联立,消去y得1?4k3, 52??x2?2464kx??0, 525???,设P?x1,y1?,Q?x2,y2?,
24k?64x?x?,, 125?20k225?100k2uuuruuur由已知EP??x1,y1?1?,EQ??x2,y2?1?,则
则x1?x2?uuuruuur8??8??EP?EQ?x1x2??y1?1??y2?1??x1x2??kx1???kx2??
5??5??864??1?k2?x1x2?k?x1?x2??
525?64824k64??1?k2???k?? 2225?100k55?20k25?64?64k2?192k2?64?256k2 ?225?100k?0,
故直线EP?EQ.
2|EP|2?|EQ|2?|PQ|2??1?k2???x1?x2??4x1x2?
??222641?k25k?4??????24k?64??2??1?k????4?? 2?2?225?20k25?100k????251?4k?????64?4?29k2?25k4?25?1?4k22?,
令1?4k2?t,则
2?t?1?t?1??64?4?29??25????2?? 444?27?66t?25t????????2|PQ|??25t225t224??133?1764?????27?????, 25?t2727?????由于t?1?4k2?1,0?1?1, t4?|PQ|2?256. 2522所以,|EP|?|EQ|的取值范围为?4,【点睛】
??256??. 25?此题考查轨迹问题,椭圆和直线相交,注意坐标表示向量进行转化的处理技巧,属于较难题目.