??1?m??1?m??1?m??1?m?2 ?1???m??????????1?m，
2222????????221?m?5215?2?1?1?m??则PQ2=PE2+EQ2=?= ········· 8分 ?m??m?2m?m???????2?222?5?10?2??2110∵0＜m＜1，∴当m= 时，PQ2 取得最小值，PQ 取得最小值， ··············· 9分
510102210∴当m= ，即Q 点的坐标为（?，0 ）时，PQ 取得最小值，
5510322此时二次函数解析式为y?x?x? ··························································· 10分
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28. （本题满分10分）
（1）连接OC，设OH＝a，∵∠CPB=45°，
222∴∠CQF=∠PQO=45°，∴FC=FQ， ································································ 1分 设FC=FQ=a，则OF=a+2，
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在Rt△OCF中，FC2?OF2?OC2?a2?(a?2)2?2a2?4a?4?32 ··················· 2分
∴2a2?4a?5
∴S矩形CEGF?2CFFO?2a(a?2)?5 ····························································· 3分 （2）不变． ································································································· 4分
∵AB垂直平分CE，∴PC=PE，且∠CPB=∠EPH=45°， ∴PE⊥CD， ∴PD2?PC2?PD2?PE2?DE2， ································································ 5分 ∵∠PCH=45°，∴∠DOE=90° ∴DE2?DO2?EO2?32?32?18 ∴PD2?PC2?18 ······································· 6分
（3）当点P在直径AB上时， 11PDPE?PDPC?3 PDE22∴PDPC?6 由（2）得PD2?PC2?18 ······················································ 7分
如图1，S?∴CD2?(PD?PC)2?18?12?30 ∴CD?30 ············································· 8分
如图2，当点P在AB延长线上，同理可得：CD2?(PC?PD)2?18?12?6 ·········· 9分 ∴CD?6 综上所述：CD?30或6 ························································ 10分
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