如图，设D(x0，kx0)，E(x1，kx1)，F(x2，kx2)，其中x1?x2， 且x1，x2满足方程(1?4k2)x2?4， 故x2??x1?y B O E F D A x 21?4k2．①
????????1510由ED?6DF知x0?x1?6(x2?x0)，得x0?(6x2?x1)?x2?；
27771?4k由D在AB上知x0?2kx0?2，得x0?所以
2． 1?2k210， ?21?2k71?4k2化简得24k?25k?6?0，
23或k?． ············································································································ 6分 38（Ⅱ）解法一：根据点到直线的距离公式和①式知，点E，F到AB的距离分别为
解得k?h1?x1?2kx1?25x2?2kx2?25?2(1?2k?1?4k2)5(1?4k)2，
h2??2(1?2k?1?4k2)5(1?4k)2． ····································································· 9分
又AB?22?1?5，所以四边形AEBF的面积为
S?1AB(h1?h2) 214(1?2k) ??5?225(1?4k)?2(1?2k)1?4k2
1?4k2?4k ?221?4k≤22，
当2k?1，即当k?1时，上式取等号．所以S的最大值为22． ······························· 12分 2解法二：由题设，BO?1，AO?2．
设y1?kx1，y2?kx2，由①得x2?0，y2??y1?0， 故四边形AEBF的面积为
S?S△BEF?S△AEF
····························································································································· 9分 ?x2?2y2 ·
?(x2?2y2)2 22?x2?4y2?4x2y2 22≤2(x2?4y2)
?22，
当x2?2y2时，上式取等号．所以S的最大值为22． ·················································· 12分