【答案】
分析:(1)連接AC,由Rt△AOC∽R(shí)t△COB?
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/0.png)
,求得OB的長(zhǎng),即可得出確定B點(diǎn)坐標(biāo),進(jìn)而可根據(jù)B、C坐標(biāo)用待定系數(shù)法求得BC直線的解析式.
(2)根據(jù)圓心的坐標(biāo)及圓的半徑不難得出E、F的坐標(biāo).根據(jù)拋物線和圓的對(duì)稱性可知:拋物線頂點(diǎn)和圓心的橫坐標(biāo)必相等,據(jù)此可根據(jù)直線BC的解析式求出拋物線的頂點(diǎn)坐標(biāo).然后根據(jù)E、F及頂點(diǎn)坐標(biāo)求出拋物線的解析式.
(3)在(1)中已經(jīng)求得C點(diǎn)坐標(biāo),將C點(diǎn)坐標(biāo)代入拋物線的解析式中進(jìn)行判斷即可.
(4)在(1)中已經(jīng)求得∠OAC=60°,∠OCA=30°,如果連接CF,那么∠CFE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/1.png)
∠OAC=30°,由于E、F同在拋物線上,因此連接CE后,三角形CEF就與三角形OAC相似.那么C、E、F就是符合條件的點(diǎn).而根據(jù)拋物線的對(duì)稱性可知,C點(diǎn)關(guān)于拋物線對(duì)稱軸的對(duì)稱點(diǎn)和E、F組成的直角三角形也應(yīng)該符合條件.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/images2.png)
解:(1)方法一:
連接AC,則AC⊥BC.
∵OA=2,AC=4,
∴OC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/2.png)
.
又∵Rt△AOC∽R(shí)t△COB,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/3.png)
.
∴OB=6.
∴點(diǎn)C坐標(biāo)為(0,2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/4.png)
),點(diǎn)B坐標(biāo)為(-6,0).
設(shè)直線BC的解析式為y=kx+b,
可求得直線BC的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/5.png)
x+2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/6.png)
.
方法二:
連接AC,則AC⊥BC.
∵OA=2,AC=4,
∴∠ACO=30°,∠CAO=60°.
∴∠CBA=30°.
∴AB=2AC=8.
∴OB=AB-AO=6.
以下同證法一.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/images8.png)
(2)由題意得,⊙A與x軸的交點(diǎn)分別為E(-2,0)、F(6,0),拋物線的對(duì)稱軸過(guò)點(diǎn)A為直線x=2.
∵拋物線的頂點(diǎn)在直線BC上,
∴拋物線頂點(diǎn)坐標(biāo)為(2,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/7.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/8.png)
).
設(shè)拋物線解析式為y=a(x-2)
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/9.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/10.png)
,
∵拋物線過(guò)點(diǎn)E(-2,0),
∴0=a(-2-2)
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/11.png)
,
解得a=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/12.png)
.
∴拋物線的解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/13.png)
(x-2)
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/14.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/15.png)
,
即y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/16.png)
x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/17.png)
x+2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/18.png)
.
(3)點(diǎn)C在拋物線上.因?yàn)閽佄锞€與y軸的交點(diǎn)坐標(biāo)為(0,2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/19.png)
),如圖.
(4)存在,這三點(diǎn)分別是E、C、F與E、C′、F,C′的坐標(biāo)為(4,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901421564/SYS201311031034249014215013_DA/20.png)
).
即△ECF∽△AOC、△EC′F∽△AOC,如圖.
點(diǎn)評(píng):本題考查了圓的相關(guān)知識(shí)、二次函數(shù)解析式的確定、相似三角形的判定和性質(zhì)等知識(shí)點(diǎn).