NANB??x1?m??x2?m??y1y2?m2?m?x1?x2??k2?x1?1??x2?1?
??1?k2?x1x2??m?k2??x1?x2??k2?m2?1?k??4k ?22?12??4k2?34k2?3?4m2?8m?5?k2?3m2?124k2?3?m?k?8k?22?k2?m2
4m2?8m?5411??m?如果要上式为定值，则必须有
3m2?1238验证当直线l斜率不存在时，也符合。 故存在点N?135?11? 9分 ,0?满足NANB??64?8?x2c312、（1）解：由2b?2,??b?1,a?2，所以椭圆方程为：?y2?1
4a2（2）解：由已知设直线l方程为：x?ky?m，设A(x1,y1),B(x2,y2),C(x3,y3),D(x4,y4)
??1?0?x?ky?m?联立?2，消x得：y2?2ky?2m?0，∴?y1?y2?2k
?y?2x?yy??2m?122y12y2?y1y2??1?y1y2??2，∴m?1 由OA?OB??1，∴x1x2?y1y2??1?4?2?0??x?ky?1?22联立?2，消得：，∴(k?4)y?2ky?3?0x?2k ?2y3?y4?2?x?4y?4?0?k?4?由|AC|?|BD|?AB中点与CD中点重合， ∴y1?y2?y3?y4?2k??2k?k?0 2k?4所以存在直线l方程为：x?1.
13、解：（1）当 P点在 x 轴上时，P?2,0?,PA:y??2(x?2) 2
?2y??(x?2)?x2??11?222??2??x?2x?1?0,??0?a?2,椭圆方程为?y2?1?22?a2? ?x?y2?1??a2（2）设切线为y?kx?m,设P?2,y0?,A?x1,y1?,则
?y?kx?m22222???0?m?2k?1, ?1?2kx?4kmx?2m?2?0???22?x?2y?2?0且x1??2kmm,y?,y0?2k?m, 1221?2k1?2ky02?4,PA直线为y?则PA?则
yx?2y1y0x?,A到直线 PO 距离d?01,
22y0?4S?POA111?2km2m1?2k2?km2?PAd?y0x1?2y1??2k?m???m?k?m?k?1?2k2221?2k21?2k21?2k22??S?k??1?2k2?k2?2Sk?S2?1?0??8S2?4?0?S?
22,此时k??. 22