（3）解：??PBE?90°，?ABC?60° ??A??CBE?30° ?AC?1由旋转的性质可知
?ACP??BCE ??APC113BC ······· (8′)
∽?BEC ·········································· (9′)
?AP:BE?AC:BC 设AP?x ?BE?1133x ····················································· (10′)
在Rt?ABC中，?A?30° ?AB?2BC?2 ?S?PBE1?12?33x(2?x)??3636x?233x??36(x?1)?236 ······························ (11′)
?x?1时 S?P1BE(max)? ························································································· (12′)
tan?DBN?DNBN?1224．解：（1）过D作DN?AB于N,
(2′) ?D（2，3），?BN?6 ?ON?2，?OB?4 ?B（-4，0） ································ 把B（-4，0），D（2，3）代入y?ax2?抛物线的解析式为y??bx?2 得a?12，b?32
12x?232··········································································· (3′) x?2
12m?2（2）过M作MH?AB于H，设M(m,S四边形BMCA?S?32···································· (4′) m?2)(m?0)
?BHM?S梯形MCOH?S12?OAC ···································· (5′)
12?1?2
12(m?4)?MH?(MH?2)(?m)???m?4m?5??(m?2)?9
22?当m??2时，S有最大值9 ················································ (7′)
（3）如右图
设AC所在直线的解析式为y?kx?2
?A（1，0） ?k?2?0?ACk?2
所在直线的解析式为y?2x?2 ·········································· (8′)
设直线AC与HM交于F，?F（-2，-6）
?AF?3?6?3522 ································································ (9′)
设?Q与直线AC相切于P 则QP?AF ································ (10′)
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设Q（-2，n），?FQ?n?6
QP?OQ?n?42 ··················· (11′) ??QPF∽?AHF?QPAH?QFAF ······················ (12′)
即n?432?n?635 化简得：n2(13′) ?3n?4?0 ?n??1或n?4 ································
?满足条件的点Q存在，其坐标为Q（?2，?1）或（?2，4）
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