【答案】
分析:(1)利用等腰梯形ABCD的面積為8求得點(diǎn)D和點(diǎn)A的坐標(biāo),然后利用待定系數(shù)法求得二次函數(shù)的解析式即可;
(2)當(dāng)點(diǎn)O在線段AD上時和當(dāng)點(diǎn)E在線段AD上時,利用△DO
1G∽△DAO求得t的值即可;
(3)分為3種情況,①旋轉(zhuǎn)后OE在拋物線上;②旋轉(zhuǎn)后OB在拋物線上;③旋轉(zhuǎn)后BE在拋物線上討論即可得到有四個不同的旋轉(zhuǎn)中心.
解答:解:(1)在等腰梯形ABCD中,S
梯形ABCD=8,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/0.png)
,
∴OD=4,
∴D(0,4),
∵tan∠DAO=4,
∴OA=1,
∴A(-1,0),
把A(-1,0)、B(2,0)、D(0,4)代入y=ax
2+bx+c得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/1.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/2.png)
.
∴y=-2x
2+2x+4.
(2)當(dāng)點(diǎn)O在線段AD上時,如圖,
BB
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/3.png)
t,B
1O
1=2,B
1H=2t,BH=t,
B
1G=2-t,O
1G=2-(2-t)=t
由△DO
1G∽△DAO得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/4.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/images5.png)
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/5.png)
,
當(dāng)點(diǎn)E在線段AD上時,如圖,
BB
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/6.png)
t,B
1H=2 t,BH=t,
∵B
1O
1=2,
∴E
1G=t,DG=4-(2t-1)=5-2t
由△DO
1G∽△DAO得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/7.png)
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/8.png)
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/9.png)
;
(3)(-2,2)(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/10.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/11.png)
) (3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/12.png)
) (-1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/13.png)
),
分為3種情況,①旋轉(zhuǎn)后OE在拋物線上;②旋轉(zhuǎn)后OB在拋物線上;③旋轉(zhuǎn)后BE在拋物線上.
1、旋轉(zhuǎn)后OE在拋物線上:
設(shè)為O′E′,則O′E′平行于x軸,拋物線y=-2x
2+2x+4=-2(x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/14.png)
)
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/15.png)
,對稱軸x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/16.png)
,
則x
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/17.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/18.png)
|OE|=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/19.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/20.png)
=0,x
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/21.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/22.png)
=1.
則兩點(diǎn)為(0,4)、(1,4).
這時分別:1)O′(0,4)、E′(1,4).
然后分兩種情況分別作OO',EE'的中垂線,其交點(diǎn)即為其旋轉(zhuǎn)中心.
∵OO′的解析式為y=2,易得,EE′的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/23.png)
x-1,則EE′的中點(diǎn)坐標(biāo)為(1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/24.png)
),
其中垂線解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/25.png)
x+b,將(1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/26.png)
)代入解析式得,b=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/27.png)
,
則解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/28.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/29.png)
,當(dāng)y=2時,x=-2.
旋轉(zhuǎn)中心坐標(biāo)為(-2,2).
2、旋轉(zhuǎn)后OB在拋物線上:
OB∥y軸,則O′B′∥x軸,但拋物線y=-2x
2+2x+8=-2(x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/30.png)
)
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/31.png)
,不成立.
3、旋轉(zhuǎn)后BE在拋物線上:
BE邊旋轉(zhuǎn)90°后所得線段B'E'與BE垂直,直線斜率k
BE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191530455162510/SYS201311011915304551625023_DA/32.png)
,則k
B'E'=-2.
設(shè)旋轉(zhuǎn)后B'E'所在直線方程為:y=-2x+m.
拋物線:y=-2x
2+2x+4,聯(lián)立,解方程,得:
(x,y)=(2,m-4-2)或 (x,y)=(2-,m-4+2)
此為兩交點(diǎn)坐標(biāo),求距離使其等于|BE|=2.有:
|BE|=2,從而有m=11,
兩點(diǎn)坐標(biāo):(3,5),(1,9).
然后分1)B′(3,5),E′(1,9);2)E′(3,5),B′(1,9)兩種情況,
分別作BB′與EE′的垂直平分線,兩者交點(diǎn)即為其旋轉(zhuǎn)中心.
綜上,同1中解法,共有5種可能性,5個旋轉(zhuǎn)中心,(-2,2),(-4,4)(5,3)(6,3)(-2,3).
點(diǎn)評:本題考查了二次函數(shù)的綜合知識,特別是二次函數(shù)的知識與旋轉(zhuǎn)、對稱、平移等知識的結(jié)合更是近幾年中考的熱點(diǎn)考題之一,難度較大.