【答案】
分析:(1)已知了A、B两点的坐标即可得出OA、OB的长,在直角三角形ACB中由于OC⊥AB,因此可用射影定理求出OC的长,即可得出C点的坐标.然后用待定系数法即可求出抛物线的解析式;
(2)本题的关键是得出D点的坐标,CD平分∠BCE,如果连接O′D,那么根据圆周角定理即可得出∠DO′B=2∠BCD=∠BCE=90°由此可得出D的坐标为(4,-5).根据B、D两点的坐标即可用待定系数法求出直线BD的解析式;
(3)本题要分两种情况进行讨论:
①过D作DP∥BC,交D点右侧的抛物线于P,此时∠PDB=∠CBD,可先用待定系数法求出直线BC的解析式,然后根据BC与DP平行,那么直线DP的斜率与直线BC的斜率相同,因此可根据D的坐标求出DP的解析式,然后联立直线DP的解析式和抛物线的解析式即可求出交点坐标,然后将不合题意的舍去即可得出符合条件的P点.
②同①的思路类似,先作与∠CBD相等的角:在O′B上取一点N,使BN=BM.可通过证△NBD≌△MDB,得出∠NDB=∠CBD,然后同①的方法一样,先求直线DN的解析式,进而可求出其与抛物线的交点即P点的坐标.
综上所述可求出符合条件的P点的值.
解答:解:(1)∵以AB为直径作⊙O′,交y轴的负半轴于点C,
∴∠OCA+∠OCB=90°,
又∵∠OCB+∠OBC=90°,
∴∠OCA=∠OBC,
又∵∠AOC=∠COB=90°,
∴△AOC∽△COB,(1分)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/0.png)
.
又∵A(-1,0),B(9,0),
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/1.png)
,
解得OC=3(负值舍去).
∴C(0,-3),
故设抛物线解析式为y=a(x+1)(x-9),
∴-3=a(0+1)(0-9),解得a=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/2.png)
,
∴二次函数的解析式为y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/3.png)
(x+1)(x-9),
即y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/4.png)
x
2-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/5.png)
x-3.(4分)
(2)∵AB为O′的直径,且A(-1,0),B(9,0),
∴OO′=4,O′(4,0),(5分)
∵点E是AC延长线上一点,∠BCE的平分线CD交⊙O′于点D,
∴∠BCD=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/6.png)
∠BCE=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/7.png)
×90°=45°,
连接O′D交BC于点M,
则∠BO′D=2∠BCD=2×45°=90°,OO′=4,O′D=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/8.png)
AB=5.
∴O′D⊥x轴
∴D(4,-5).(6分)
∴设直线BD的解析式为y=kx+b(k≠0)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/9.png)
(7分)
解得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/10.png)
∴直线BD的解析式为y=x-9.(8分)
(3)假设在抛物线上存在点P,使得∠PDB=∠CBD,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/images11.png)
解法一:设射线DP交⊙O′于点Q,则
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/11.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/12.png)
.
分两种情况(如图所示):
①∵O′(4,0),D(4,-5),B(9,0),C(0,-3).
∴把点C、D绕点O′逆时针旋转90°,使点D与点B重合,则点C与点Q
1重合,
因此,点Q
1(7,-4)符合
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/13.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/14.png)
,
∵D(4,-5),Q
1(7,-4),
∴用待定系数法可求出直线DQ
1解析式为y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/15.png)
x-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/16.png)
.(9分)
解方程组
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/17.png)
得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/18.png)
∴点P
1坐标为(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/19.png)
,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/20.png)
),坐标为(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/21.png)
,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/22.png)
)不符合题意,舍去.(10分)
②∵Q
1(7,-4),
∴点Q
1关于x轴对称的点的坐标为Q
2(7,4)也符合
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/23.png)
=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/24.png)
.
∵D(4,-5),Q
2(7,4).
∴用待定系数法可求出直线DQ
2解析式为y=3x-17.(11分)
解方程组
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/25.png)
得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/26.png)
,
即
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/27.png)
∴点P
2坐标为(14,25),坐标为(3,-8)不符合题意,舍去.(12分)
∴符合条件的点P有两个:P
1(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/28.png)
,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/29.png)
),P
2(14,25).
解法二:分两种情况(如图所示):
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/images31.png)
①当DP
1∥CB时,能使∠PDB=∠CBD.
∵B(9,0),C(0,-3).
∴用待定系数法可求出直线BC解析式为y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/30.png)
x-3.
又∵DP
1∥CB,
∴设直线DP
1的解析式为y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/31.png)
x+n.
把D(4,-5)代入可求n=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/32.png)
,
∴直线DP
1解析式为y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/33.png)
x-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/34.png)
.(9分)
解方程组
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/35.png)
得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/36.png)
∴点P
1坐标为(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/37.png)
,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/38.png)
)或(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/39.png)
,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/40.png)
)(不符合题意舍去).(10分)
②在线段O′B上取一点N,使BN=DM时,得△NBD≌△MDB(SAS),
∴∠NDB=∠CBD.
由①知,直线BC解析式为y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/41.png)
x-3.
取x=4,得y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/42.png)
,
∴M(4,-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/43.png)
),
∴O′N=O′M=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/44.png)
,
∴N(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/45.png)
,0),
又∵D(4,-5),
∴直线DN解析式为y=3x-17.(11分)
解方程组
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/46.png)
得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/47.png)
,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/48.png)
∴点P
2坐标为(14,25),坐标为(3,-8)不符合题意,舍去.(12分)
∴符合条件的点P有两个:P
1(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/49.png)
,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/50.png)
),P
2(14,25).
解法三:分两种情况(如图所示):
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/images53.png)
①求点P
1坐标同解法二.(10分)
②过C点作BD的平行线,交圆O′于G,
此时,∠GDB=∠GCB=∠CBD.
由(2)题知直线BD的解析式为y=x-9,
又∵C(0,-3)
∴可求得CG的解析式为y=x-3,
设G(m,m-3),作GH⊥x轴交于x轴与H,
连接O′G,在Rt△O′GH中,利用勾股定理可得,m=7,
由D(4,-5)与G(7,4)可得,
DG的解析式为y=3x-17,(11分)
解方程组
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/51.png)
得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/52.png)
,
即
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/53.png)
∴点P
2坐标为(14,25),坐标为(3,-8)不符合题意舍去.(12分)
∴符合条件的点P有两个:P
1(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/54.png)
,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/55.png)
),P
2(14,25).
说明:本题解法较多,如有不同的正确解法,请按此步骤给分.
解:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/images59.png)
过B作BM⊥CD于M,
B(9,0),C(0,-3),由勾股定理得:BC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/56.png)
=3
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/57.png)
,
∵∠BCD=45°,
∴BM=CM,
由勾股定理得:BM=3
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/58.png)
,
∵△PCD的面积是△BCD面积的三分之一,
∴根据△CDB和△CDP有一条公共边CD,得出P到CD的高是3
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/59.png)
÷3=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/60.png)
,
根据C(0,-3),D(4,-5)的坐标求出直线CD的解析式是y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/61.png)
x-3,
把直线CD向上平移
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/62.png)
单位得出直线y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/63.png)
x-3+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/64.png)
,把直线CD向下平移
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/65.png)
单位得出直线y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/66.png)
x-3-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/67.png)
,
则
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/68.png)
,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/69.png)
,
解得:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/70.png)
(因为此点不在直线BC下方舍去),
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/71.png)
,(因为此点不在直线BC下方舍去),
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/72.png)
,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/73.png)
.
即P的坐标是(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/74.png)
,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/75.png)
)或(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/76.png)
,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021230646688027596/SYS201310212306466880275023_DA/77.png)
).
点评:本题着重考查了待定系数法求二次函数解析式、三角形相似及全等、探究角相等的构成情况等知识点,综合性强,考查学生分类讨论,数形结合的数学思想方法.