【答案】
分析:(1)由∠ACB=90°,AC=BC,OA=1,OC=4,可得A(-1,0)B(4,5),然后利用待定系數(shù)法即可求得b,c的值;
(2)由直線AB經(jīng)過(guò)點(diǎn)A(-1,0),B(4,5),即可求得直線AB的解析式,又由二次函數(shù)y=x
2-2x-3,設(shè)點(diǎn)E(t,t+1),則可得點(diǎn)F的坐標(biāo),則可求得EF的最大值,求得點(diǎn)E的坐標(biāo);
(3)①順次連接點(diǎn)E、B、F、D得四邊形EBFD,可求出點(diǎn)F的坐標(biāo)(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/0.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/1.png)
),點(diǎn)D的坐標(biāo)為(1,-4)由S
四邊形EBFD=S
△BEF+S
△DEF即可求得;
②過(guò)點(diǎn)E作a⊥EF交拋物線于點(diǎn)P,設(shè)點(diǎn)P(m,m
2-2m-3),可得m
2-2m-3=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/2.png)
,即可求得點(diǎn)P的坐標(biāo),又由過(guò)點(diǎn)F作b⊥EF交拋物線于P
3,設(shè)P
3(n,n
2-2n-3),可得n
2-2n-2=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/3.png)
,求得點(diǎn)P的坐標(biāo),則可得使△EFP是以EF為直角邊的直角三角形的P的坐標(biāo).
解答:解:(1)由已知得:A(-1,0),B(4,5),
∵二次函數(shù)y=x
2+bx+c的圖象經(jīng)過(guò)點(diǎn)A(-1,0),B(4,5),
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/4.png)
,
解得:b=-2,c=-3;
(2)如圖:∵直線AB經(jīng)過(guò)點(diǎn)A(-1,0),B(4,5),
∴直線AB的解析式為:y=x+1,
∵二次函數(shù)y=x
2-2x-3,
∴設(shè)點(diǎn)E(t,t+1),則F(t,t
2-2t-3),
∴EF=(t+1)-(t
2-2t-3)=-(t-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/5.png)
)
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/6.png)
,
∴當(dāng)t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/7.png)
時(shí),EF的最大值為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/8.png)
,
∴點(diǎn)E的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/9.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/10.png)
);
(3)①如圖:順次連接點(diǎn)E、B、F、D得四邊形EBFD.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/images11.png)
可求出點(diǎn)F的坐標(biāo)(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/11.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/12.png)
),點(diǎn)D的坐標(biāo)為(1,-4)
S
四邊形EBFD=S
△BEF+S
△DEF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/13.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/14.png)
×(4-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/15.png)
)+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/16.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/17.png)
×(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/18.png)
-1)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/19.png)
;
②如圖:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/images21.png)
�。┻^(guò)點(diǎn)E作a⊥EF交拋物線于點(diǎn)P,設(shè)點(diǎn)P(m,m
2-2m-3)
則有:m
2-2m-3=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/20.png)
,
解得:m
1=1+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/21.png)
,m
2=1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/22.png)
,
∴P
1(1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/23.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/24.png)
),P
2(1+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/25.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/26.png)
),
ⅱ)過(guò)點(diǎn)F作b⊥EF交拋物線于P
3,設(shè)P
3(n,n
2-2n-3)
則有:n
2-2n-3=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/27.png)
,
解得:n
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/28.png)
,n
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/29.png)
(與點(diǎn)F重合,舍去),
∴P
3(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/30.png)
,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/31.png)
),
綜上所述:所有點(diǎn)P的坐標(biāo):P
1(1+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/32.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/33.png)
),P
2(1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/34.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/35.png)
),P
3(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/36.png)
,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190318892795212/SYS201311011903188927952027_DA/37.png)
)能使△EFP組成以EF為直角邊的直角三角形.
點(diǎn)評(píng):此題考查了待定系數(shù)法求二次函數(shù)的解析式,四邊形與三角形面積問(wèn)題以及直角三角形的性質(zhì)等知識(shí).此題綜合性很強(qiáng),解題的關(guān)鍵是注意方程思想與數(shù)形結(jié)合思想的應(yīng)用.