设经过D，P，C三点的抛物线表达式为y?ax2?bx?c， 1?a?，?4?c?7，?11??则?49a?7b?c?0， ∴?b??，
4??64a?8b?c?1．??c?7．??∴所求抛物线的表达式为：y?14x?2114······················································ 10分 x?7． ·
（说明：求出一条抛物线表达式给3分，求出两条抛物线表达式给4分）
25．解：（1）答案不唯一，如图①、②（只要满足题意，画对一个图形给2分，画对两个给3分）
O B A D A B
O C
C
（第25题答案图①） （第25题答案图②）
·····································································································································3分 （2）过点A，B分别作CD的垂线，垂足分别为M，N．
∵S△ACD?S△BCD?1212CD·AM?1212CD·AE·sin?，
CD·BN?CD·BE·sin?． ·······································································5分
∴S四边形ACBD?S△ACD?S△BCD
?12121212CD·AE·sin??12CD·BE·sin?
???CD·(AE?BE·)sin? CD·AB·sin?
msin?． ·······································7分
2A ED MN? O C B
（第25题答案图③）
（3）存在．分两种情况说明如下： ············································································· 8分 ①当AB与CD相交时， 由（2）及AB?CD?2R知S四边形ACBD?12····················9分 AB·CD·sin??Rsin?． ·
A 2 2②当AB与CD不相交时，如图④ ∵AB?CD?2R，OC?OD?OA?OB?R，
D H E C ∴?AOB??COD?90°，
而S四边形ABCD?SRt△AOB?SRt△OCD?S△AOD?S△BOC
?R?S△AOD?S△BOC2B O 3 1 （第25题答案图④）
．····························································································· 10分
延长BO交?O于点E，连接EC，则?1??3??2??3?90°．
∴?1??2．
∴△AOD≌△COE．
∴S△AOD?S△OCE．
∴S△AOD?S△BOC?S△OCE?S△BOC?S△BCE．
过点C作CH?BE，垂足为H， 则S△BCE?12BE·CH?R·CH．
2∴当CH?R时，S△BCE取最大值R． ······································································ 11分
综合①、②可知，当?1??2?90°，即四边形ABCD是边长为2R的正方形时， S四边形ABCD?R?R?2R为最大值． ······································································· 12分
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