【答案】
分析:(1)欲畫△ECD關(guān)于邊CD所在直線為對稱軸的對稱圖形△E
1CD,由CD不變,知關(guān)鍵是確定E
1點(diǎn),可過點(diǎn)E作對稱軸CD的垂線,垂足為F,延長EF到E
1,使E
1F=EF.則點(diǎn)E
1就是點(diǎn)E關(guān)于CD所在直線的對稱點(diǎn);
(2)由(1)求得E
1坐標(biāo),再求拋物線的函數(shù)表達(dá)式,可通過待定系數(shù)法,利用已知條件求解;
(3)問題較難,根據(jù)兩個三角形相似的條件,需要分情況討論P(yáng)在不同位置時的情況.
解答:解:(1)過點(diǎn)E作EE
1⊥CD交BC于F點(diǎn),交x軸于E
1點(diǎn),
則E
1點(diǎn)為E的對稱點(diǎn).連接DE
1、CE
1,則△CE
1D為所畫的三角形,
∵△CED∽△OEA,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/0.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/1.png)
,
∵EF、EE
1分別是△CED、△OEA的對應(yīng)高,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/images2.png)
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/3.png)
,
∴EF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/4.png)
EE
1,
∴F是EE
1的中點(diǎn),
∴E點(diǎn)關(guān)于CD的對稱點(diǎn)是E
1點(diǎn),△CE
1D為△CED關(guān)于CD的對稱圖形,
在Rt△EOE
1,OE
1=cos60°×EO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/5.png)
×8=4,
∴E
1點(diǎn)的坐標(biāo)為(4,0);
(2)∵平行四邊形OABC的高為h=sin60°×4=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/6.png)
,
過C作CG⊥OA于G,則OG=2,
∴C、B點(diǎn)的坐標(biāo)分別為(2,2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/7.png)
),(8,2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/8.png)
),
∵拋物線過C、B兩點(diǎn),且CB∥x軸,C、B兩點(diǎn)關(guān)于拋物線的對稱軸對稱,
∴拋物線的對稱軸方程為x=5,
又∵拋物線經(jīng)過E
1(4,0),
則拋物線與x軸的另一個交點(diǎn)為A(6,0),
∴可設(shè)拋物線為y=a(x-4)(x-6),
∵點(diǎn)C(2,2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/9.png)
)在拋物線上,
∴2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/10.png)
=a(2-4)(2-6),
解得a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/11.png)
,
∴y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/12.png)
(x-4)(x-6)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/13.png)
x
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/14.png)
x+6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/15.png)
;
(3)根據(jù)兩個三角形相似的條件,由于在△ECD中,∠ECD=60°,
若△BCP與△ECD相似,則△BCP中必有一個角為60°,
下面進(jìn)行分類討論:
①當(dāng)P點(diǎn)直線CB的上方時,由于△PCB中,∠CBP>90°或∠BCP>90°,
∴△PCB為鈍角三角形,
又∵△ECD為銳角三角形,
∴△ECD與△CPB不相似.
從而知在直線CB上方的拋物線上不存在點(diǎn)P使△CPB與△ECD相似;
②當(dāng)P點(diǎn)在直線CB上時,點(diǎn)P與C點(diǎn)或B點(diǎn)重合,不能構(gòu)成三角形,
∴在直線CB上不存在滿足條件的P點(diǎn);
③當(dāng)P點(diǎn)在直線CB的下方時,若∠BCP=60°,則P點(diǎn)與E
1點(diǎn)重合,
此時,∠ECD=∠BCE
1,而
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/16.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/17.png)
,
∴△BCE與△ECD不相似,
若∠CBP=60°,則P點(diǎn)與A點(diǎn)重合,
根據(jù)拋物線的對稱性,同理可證△BCA與△CED不相似,
若∠CPB=60°,假設(shè)拋物線上存在點(diǎn)P使△CPB與△ECD相似,
∴EF=sin60°×4=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/18.png)
,F(xiàn)D=1,
∴ED=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/19.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/20.png)
,
設(shè)△ECD的邊DE上的高為h
1,則有
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/21.png)
h
1×ED=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/22.png)
EF×CD,
∴h
1=EF×CD÷ED=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/23.png)
×3÷
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/24.png)
=6
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/25.png)
÷
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/26.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/27.png)
,
設(shè)△CPB的邊BC上的高為h
2,△CPB與△ECD相似,
∵
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/28.png)
,
解得h
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/29.png)
×h
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/30.png)
×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/31.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/32.png)
,
∵拋物線的頂點(diǎn)坐標(biāo)為(5,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/33.png)
),
∴拋物線的頂點(diǎn)到直線BC的距離d=|-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/34.png)
|+2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/35.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193100598195416/SYS201311011931005981954024_DA/36.png)
,
∵h(yuǎn)
2>d,
∴所求P點(diǎn)到直線BC的距離大于拋物線的頂點(diǎn)到直線BC的距離,
從而使△CPB與△ECD相似的點(diǎn)P不會在拋物線上,
∴在直線CB下方不存在拋物線上的點(diǎn)P使△CPB與△ECD相似.
綜上所述,拋物線上不存在點(diǎn)P使點(diǎn)P、B、C為頂點(diǎn)的三角形與△ECD相似.
點(diǎn)評:(1)考查的是作簡單平面圖形軸對稱后的圖形,其依據(jù)是軸對稱的性質(zhì).
基本作法:①先確定圖形的關(guān)鍵點(diǎn);
②利用軸對稱性質(zhì)作出關(guān)鍵點(diǎn)的對稱點(diǎn);
③按原圖形中的方式順次連接對稱點(diǎn).
(2)用待定系數(shù)法求函數(shù)的解析式時要靈活地根據(jù)已知條件選擇配方法和公式法.
(3)是一道難度較大的二次函數(shù)題,綜合考查了三角形相似的性質(zhì),需注意分類討論,全面考慮點(diǎn)P所在位置的各種情況.