设直线l交椭圆于A?x1,y1?,B?x2,y2?
?x?my?1?由?x2y2??3m2?4?y2?6my?9?0
?1??3?4∴???6m??363m2?4?144m2?1?0
2
????∴y1?y2??
6m9 ?????6分 ,y?y??12223m?43m?4?????????1?又由MA??1AF??x1,y1????1?1?x1,?y1?
m????1??1?1 my11 my2
同理?2??1?
∴?1??2??2?1?11???? ?m?y1y2???3m2?4?2my1?y2116m?????2????∵ ??y1y2y1y2933m?4??1?11?12m8?????2????∴?1??2??2?? ?????9分 m?y1y2?m33?所以，当m变化时， ?1??2的值为定值?8； ?????10分 3 （Ⅲ）证明：由（Ⅱ）知A?x1,y1?,B?x2,y2?，∴D(4,y1),E(4,y2)
方法1）∵lAE:y?y2?y2?y1??x?4?
4?x1当x?y?y1?3?2?4?x1??y2?3?y2?y1?5?????时，y?y2?2
??4?x1?2?24?x12?2?4?my1?1??y2?3?y2?y1?3?y2?y1??2my1y2?
2?4?x1?2?4?x1?3??6m?9?2m?3m2?43m2?4?0
2?4?x1??
∴点N?,0?在直线lAE上，
?5?2??同理可证，点N?,0?也在直线lBD上；
?5?2??∴当m变化时，AE与BD相交于定点?,0??????14分 方法2）∵kEN??5?2??y24?52?2y2 3kAN?y1x1?52?y1my1?1?52?2y1
2my1?3kEN?kAN?2y22y12y?2my1?3??6y1??2 32my1?33?2my1?3?4m??9?6m?6?3m2?43m2?4?0 3?2my1?3??
4my1y2?6?y1?y2??3?2my1?3?∴kEN?kAN∴A、N、E三点共线， 同理可得B、N、D也三点共线；
∴当m变化时，AE与BD相交于定点?,0? ?????14分
?5?2??