2??ba??3?2a?1??4??24、解：（1）由题意得?9a?3b?c?0，解得?b?
3???c??2????c??2?∴此抛物线的解析式为y?224x?x?2 333分
（2）连结AC、BC.因为BC的长度一定，
所以△PBC周长最小，就是使PC?PB最小.B点关于对称轴的对称点是A点，
AC与对称轴x??1的交点即为所求的点P.
设直线AC的表达式为y?kx?b
E y 2???3k?b?0，?k??则? 解得?3 ?b??2??b??2∴此直线的表达式为y??把x??1代入得y??A P O B D x C 2x?2．……5分 3（第24题图）
44??∴P点的坐标为??1··········································· 6分 ，??·33??（3）S存在最大值 ································································································· 7分 理由：∵DE∥PC，即DE∥AC．
ODOE2?mOE?，?．即 OCOA2333∴OE?3?m，AE?3，OE?m
22∴△OED∽△OAC．∴方法一：
连结OPS?S四边形PDOE?S△OED?S△POE?S△POD?S△OED =
1?3?411?3???3?m?????2?m??1???3?m???2?m? 2?2?322?2?323m?m ····································································································· 8分 423333∵??0，∴当m?1时，S最大???? ·············································· 9分
4424=?方法二：
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S?S△OAC?S△OED?S△AEP?S△PCD
=
11?3?1341?3?2???3?m???2?m???m???m?1 22?2?2232323332m?m???m?1?? ····································································· 8分 424433∵??0，∴当m?1时，S最大? ······························································· 9分
=?44073】(1)∵∠A、∠C所对的圆弧相同，∴∠A＝∠C． ∴Rt△APD∽Rt△CPB，∴
APCP?PDPB，∴PA·
PB＝PC·PD；………………………3分
(2)∵F为BC的中点，△BPC为Rt△，∴FP＝FC，∴∠C＝∠CPF． 又∠C＝∠A，∠DPE＝∠CPF，∴∠A＝∠DPE．∵∠A＋∠D＝90°， ∴∠DPE＋∠D＝90°．∴EF⊥AD．
(3)作OM⊥AB于M，ON⊥CD于N，同垂径定理： ∴OM2＝(25)2－42＝4，ON2＝(25)2－32＝11 又易证四边形MONP是矩形， ∴OP＝OM2?ON2?15
【
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