【答案】
分析:(1)設(shè)直線AB的函數(shù)解析式為y=kx+4,把(4,0)代入即可;
(2)①先證出△BDO≌△COD,得出∠BDO=∠CDO,再根據(jù)∠CDO=∠ADP,即可得出∠BDE=∠ADP,
②先連結(jié)PE,根據(jù)∠ADP=∠DEP+∠DPE,∠BDE=∠ABD+∠OAB,∠ADP=∠BDE,∠DEP=∠ABD,得出∠DPE=∠OAB,再證出∠DFE=∠DPE=45°,最后根據(jù)∠DEF=90°,得出△DEF是等腰直角三角形,從而求出DF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/0.png)
DE,即y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/1.png)
x;
(3)當(dāng)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/2.png)
=2時(shí),過(guò)點(diǎn)F作FH⊥OB于點(diǎn)H,則∠DBO=∠BFH,再證出△BOD∽△FHB,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/3.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/4.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/5.png)
=2,得出FH=2,OD=2BH,再根據(jù)∠FHO=∠EOH=∠OEF=90°,得出四邊形OEFH是矩形,OE=FH=2,EF=OH=4-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/6.png)
OD,根據(jù)DE=EF,求出OD的長(zhǎng),從而得出直線CD的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/7.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/8.png)
,最后根據(jù)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/9.png)
求出點(diǎn)P的坐標(biāo)即可;
當(dāng)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/10.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/11.png)
時(shí),連結(jié)EB,先證出△DEF是等腰直角三角形,過(guò)點(diǎn)F作FG⊥OB于點(diǎn)G,同理可得△BOD∽△FGB,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/12.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/13.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/14.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/15.png)
,得出FG=8,OD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/16.png)
BG,再證出四邊形OEFG是矩形,求出OD的值,再求出直線CD的解析式,最后根據(jù)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/17.png)
即可求出點(diǎn)P的坐標(biāo).
解答:解:(1)設(shè)直線AB的函數(shù)解析式為y=kx+4,
代入(4,0)得:4k+4=0,
解得:k=-1,
則直線AB的函數(shù)解析式為y=-x+4;
(2)①由已知得:
OB=OC,∠BOD=∠COD=90°,
又∵OD=OD,
∴△BDO≌△CDO,
∴∠BDO=∠CDO,
∵∠CDO=∠ADP,
∴∠BDE=∠ADP,
②連結(jié)PE,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/images18.png)
∵∠ADP是△DPE的一個(gè)外角,
∴∠ADP=∠DEP+∠DPE,
∵∠BDE是△ABD的一個(gè)外角,
∴∠BDE=∠ABD+∠OAB,
∵∠ADP=∠BDE,∠DEP=∠ABD,
∴∠DPE=∠OAB,
∵OA=OB=4,∠AOB=90°,
∴∠OAB=45°,
∴∠DPE=45°,
∴∠DFE=∠DPE=45°,
∵DF是⊙Q的直徑,
∴∠DEF=90°,
∴△DEF是等腰直角三角形,
∴DF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/18.png)
DE,即y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/19.png)
x;
(3)當(dāng)BD:BF=2:1時(shí),
過(guò)點(diǎn)F作FH⊥OB于點(diǎn)H,
∵∠DBO+∠OBF=90°,∠OBF+∠BFH=90°,
∴∠DBO=∠BFH,
又∵∠DOB=∠BHF=90°,
∴△BOD∽△FHB,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/20.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/21.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/22.png)
=2,
∴FH=2,OD=2BH,
∵∠FHO=∠EOH=∠OEF=90°,
∴四邊形OEFH是矩形,
∴OE=FH=2,
∴EF=OH=4-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/23.png)
OD,
∵DE=EF,
∴2+OD=4-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/24.png)
OD,
解得:OD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/25.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/images27.png)
∴點(diǎn)D的坐標(biāo)為(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/26.png)
),
∴直線CD的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/27.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/28.png)
,
由
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/29.png)
得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/30.png)
,
則點(diǎn)P的坐標(biāo)為(2,2);
當(dāng)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/31.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/32.png)
時(shí),
連結(jié)EB,同(2)①可得:∠ADB=∠EDP,
而∠ADB=∠DEB+∠DBE,∠EDP=∠DAP+∠DPA,
∵∠DEB=∠DPA,
∴∠DBE=∠DAP=45°,
∴△DEF是等腰直角三角形,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/images35.png)
過(guò)點(diǎn)F作FG⊥OB于點(diǎn)G,
同理可得:△BOD∽△FGB,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/33.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/34.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/35.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/36.png)
,
∴FG=8,OD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/37.png)
BG,
∵∠FGO=∠GOE=∠OEF=90°,
∴四邊形OEFG是矩形,
∴OE=FG=8,
∴EF=OG=4+2OD,
∵DE=EF,
∴8-OD=4+2OD,
OD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/38.png)
,
∴點(diǎn)D的坐標(biāo)為(0,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/39.png)
),
直線CD的解析式為:y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/40.png)
x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/41.png)
,
由
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/42.png)
得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193239856073382/SYS201311011932398560733025_DA/43.png)
,
∴點(diǎn)P的坐標(biāo)為(8,-4),
綜上所述,點(diǎn)P的坐標(biāo)為(2,2)或(8,-4).
點(diǎn)評(píng):此題考查了一次函數(shù)的綜合,用到的知識(shí)點(diǎn)是一次函數(shù)、矩形的性質(zhì)、圓的性質(zhì),關(guān)鍵是綜合運(yùn)用有關(guān)知識(shí)作出輔助線,列出方程組.