解得a?4?4?8或a?4，因为f(0)?8，f(2)?2，f???. 3?3?9 (9分)
所以当a?488，f(a)取得最小值.，因此面积S1?S2的最小值 399 22.（1）设直
线AB方程为:x?ky?m,抛物线方程为:y2?2px(p?0),
?y2?2px由 ? 得, y2?2pky?2pm?0,
?x?ky?m设A(x1,y1),B(x2,y2),则有??y1?y2?2pk,
yy??2pm?12由题意, |?2pm|?2m?2p?2, 故所求抛物线方程为:y2?2x; (2)S?AOB?11|OM||y1?y2|?m4k2?8m?m2m 22(3)cos?AOB?OA?OB|OA|?|OB|?x1x2?y1y22(x12?2x1)(x2?2x2)?1??,
22m4k2?8m3(m2?2m)?
?0?m?22 ?0?m??223?3m?20m?12?4k?0