∵△DEF为直角三角形，∴要使△DEF为等腰三角形，只能DE＝EF 又DE ＝CD ＋CE ＝m＋x，EF ＝BE ＋BF ＝( 8－x)＋y＝( 8－x)＋
22222222222
144 m22144144，即m＋16x－64－＝0 m2m2242144当x＝2时，m－32－2＝0，即m－32m－144＝0
m∴m＋x＝( 8－x)＋
222
解得m＝36或m＝－4（舍去）
22
∵m＞0，∴m＝6 ······························································································· 10分
242144当x＝6时，m＋32－2＝0，即m＋32m－144＝0
m
解得m＝－36（舍去）或m＝4
22
∵m＞0，∴m＝2 ······························································································· 12分
28、解：（1）设直线AB的解析式为y＝px＋q
1??3＝－4p＋q?p＝－?
2 则? 解得?
?0＝2p＋q???q＝1
1∴直线AB的解析式为y＝－x＋1 ··································································· 2分
2y ∵当x＝3和x＝－3时，这条抛物线上对应点的纵坐标相等 ∴抛物线的对称轴为y轴，∴b＝0，∴y＝ax＋c 把A（－4，3）、B（2，0）代入，得：
1??3＝16a＋c?a＝?
4 ? 解得?
??0＝4a＋c??c＝－1
12
∴抛物线的解析式为y＝x－1 ························· 4分
4
2
A B O x C l （2）∵A（－4，3），∴AO＝32＋42＝5，即⊙A的半径为5
E ∵经过点C（0，－2）的直线l与x轴平行
∴直线l的解析式为y＝－2，∴点A到直线l的距离为5
∴直线l与⊙A相切······························································································ 8分 （3）把x＝－1代入y＝－
133x＋1，得y＝，∴D（－1，） 22212
m＋1 4
过点P作PH⊥直线l于H，则PH＝n＋2，即
112
又∵PO＝m2＋n2＝m2＋(m2?1)2＝m＋1
44
∴PH＝PO ············································································································ 10分 ∵DO的长度为定值，∴当PD＋PO即PD＋PH最小时，△PDO的周长最小 当D、P、H三点在一条直线上时，PD＋PH最小 ∴点P的横坐标为－1，代入抛物线的解析式，得n＝－∴P（－1，－
3 43） ················································ 12分 4此时四边形CODP的面积为： S四边形CODP＝S△PDO＋S△PCO
＝
133117×( ＋)×1＋×2×1＝················ 14分 22428y A D B O P H C l x