【答案】
分析:(1)可根據(jù)A,B的坐標(biāo),用交點(diǎn)式二次函數(shù)通式來(lái)設(shè)出拋物線的解析式,進(jìn)而可得出D,C的坐標(biāo).
(2)本題的關(guān)鍵是求出a的值.可通過(guò)相似三角形來(lái)求解,過(guò)D作DE⊥y軸于E,易知△DEC∽△COB,可通過(guò)得出的關(guān)于DE,CO,EC,OB的比例關(guān)系式,求出a的值.進(jìn)而可求出拋物線的解析式.
(3)本題要分兩種情況進(jìn)行討論.
①當(dāng)∠BDQ=90°時(shí),此時(shí)DQ是圓G的切線,設(shè)DQ交y軸于M,那么可通過(guò)求直線DM的解析式,然后聯(lián)立拋物線的解析式即可求出Q點(diǎn)的坐標(biāo).
②當(dāng)∠DBQ=90°時(shí),可過(guò)Q作x軸的垂線,設(shè)垂足為F,先設(shè)出Q點(diǎn)的坐標(biāo),然后根據(jù)相似三角形DHB和BFQ得出的關(guān)于DH,BF,BH,F(xiàn)Q的比例關(guān)系式,求出Q點(diǎn)的坐標(biāo).
③當(dāng)∠BQD=90°時(shí),顯然此時(shí)Q,C重合,因此Q點(diǎn)的坐標(biāo)即為C點(diǎn)的坐標(biāo).
綜上所述可得出符合條件的Q點(diǎn)的坐標(biāo).
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129575130505/SYS201311031011295751305000_DA/images0.png)
解:(1)設(shè)拋物線的解析式為y=a(x+1)(x-3)
則y=a(x
2-2x-3)=a(x-1)
2-4a
則點(diǎn)D的坐標(biāo)為D(1,-4a)
點(diǎn)C的坐標(biāo)為C(0,-3a)
(2)如圖①所示,過(guò)點(diǎn)D作DE⊥y軸于E,如圖①所示:
則有△DEC∽△COB
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129575130505/SYS201311031011295751305000_DA/0.png)
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129575130505/SYS201311031011295751305000_DA/1.png)
∴a
2=1a=±1
故拋物線的解析式為y=x
2-2x-3或y=-x
2+2x+3;
(3)a<0時(shí),a=-1,拋物線y=-x
2+2x+3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129575130505/SYS201311031011295751305000_DA/images3.png)
這時(shí)可以找到點(diǎn)Q,很明顯,點(diǎn)C即在拋物線上,
又在⊙G上,∠BCD=90°,這時(shí)Q與C點(diǎn)重合,點(diǎn)Q坐標(biāo)為Q(0,3).
如圖②,若∠DBQ為90°,作QF⊥y軸于F,DH⊥x軸于H
可證Rt△DHB∽R(shí)t△BFQ
有
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129575130505/SYS201311031011295751305000_DA/2.png)
則點(diǎn)Q坐標(biāo)(k,-k
2+2k+3)
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129575130505/SYS201311031011295751305000_DA/3.png)
化簡(jiǎn)為2k
2-3k-9=0
即(k-3)(2k+3)=0
解之為k=3或
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129575130505/SYS201311031011295751305000_DA/4.png)
.
由
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129575130505/SYS201311031011295751305000_DA/5.png)
得Q坐標(biāo):
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129575130505/SYS201311031011295751305000_DA/6.png)
.
若∠BDQ為90°,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129575130505/SYS201311031011295751305000_DA/images9.png)
如圖③,延長(zhǎng)DQ交y軸于M,
作DE⊥y軸于E,DH⊥x軸于H
可證明△DEM∽△DHB
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129575130505/SYS201311031011295751305000_DA/7.png)
,
則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129575130505/SYS201311031011295751305000_DA/8.png)
得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129575130505/SYS201311031011295751305000_DA/9.png)
,點(diǎn)M的坐標(biāo)為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129575130505/SYS201311031011295751305000_DA/10.png)
DM所在的直線方程為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129575130505/SYS201311031011295751305000_DA/11.png)
則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129575130505/SYS201311031011295751305000_DA/12.png)
與y=-x
2+2x+3的解為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129575130505/SYS201311031011295751305000_DA/13.png)
,
得交點(diǎn)坐標(biāo)Q為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129575130505/SYS201311031011295751305000_DA/14.png)
即滿足題意的Q點(diǎn)有三個(gè),(0,3),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101129575130505/SYS201311031011295751305000_DA/15.png)
.
點(diǎn)評(píng):本題主要考查了二次函數(shù)解析式的確定、相似三角形的判定和應(yīng)用、函數(shù)圖象交點(diǎn)等知識(shí),綜合性強(qiáng),考查學(xué)生分類討論,數(shù)形結(jié)合的數(shù)學(xué)思想方法.