【答案】
分析:(1)利用待定系數(shù)法即可求得函數(shù)的解析式;
(2)利用勾股定理求得△BCD的三邊的長,然后根據(jù)勾股定理的逆定理即可作出判斷;
(3)分p在x軸和y軸兩種情況討論,舍出P的坐標,根據(jù)相似三角形的對應邊的比相等即可求解.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/images0.png)
解:(1)設拋物線的解析式為y=ax
2+bx+c
由拋物線與y軸交于點C(0,3),可知c=3.即拋物線的解析式為y=ax
2+bx+3.
把點A(1,0)、點B(-3,0)代入,得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/0.png)
解得a=-1,b=-2
∴拋物線的解析式為y=-x
2-2x+3.
∵y=-x
2-2x+3=-(x+1)
2+4
∴頂點D的坐標為(-1,4);
(2)△BCD是直角三角形.
理由如下:解法一:過點D分別作x軸、y軸的垂線,垂足分別為E、F.
∵在Rt△BOC中,OB=3,OC=3,
∴BC
2=OB
2+OC
2=18
在Rt△CDF中,DF=1,CF=OF-OC=4-3=1,
∴CD
2=DF
2+CF
2=2
在Rt△BDE中,DE=4,BE=OB-OE=3-1=2,
∴BD
2=DE
2+BE
2=20
∴BC
2+CD
2=BD
2∴△BCD為直角三角形.
解法二:過點D作DF⊥y軸于點F.
在Rt△BOC中,∵OB=3,OC=3
∴OB=OC∴∠OCB=45°
∵在Rt△CDF中,DF=1,CF=OF-OC=4-3=1
∴DF=CF
∴∠DCF=45°
∴∠BCD=180°-∠DCF-∠OCB=90°
∴△BCD為直角三角形.
(3)①△BCD的三邊,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/1.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/3.png)
,又
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/4.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/5.png)
,故當P是原點O時,△ACP∽△DBC;
②當AC是直角邊時,若AC與CD是對應邊,設P的坐標是(0,a),則PC=3-a,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/6.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/7.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/8.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/9.png)
,解得:a=-9,則P的坐標是(0,-9),三角形ACP不是直角三角形,則△ACP∽△CBD不成立;
③當AC是直角邊,若AC與BC是對應邊時,設P的坐標是(0,b),則PC=3-b,則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/10.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/11.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/12.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/13.png)
,解得:b=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/14.png)
,故P是(0,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/15.png)
)時,則△ACP∽△CBD一定成立;
④當P在y軸上時,AC是直角邊,P一定在B的左側,設P的坐標是(d,0).
則AB=1-d,當AC與CD是對應邊時,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/16.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/17.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/18.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/19.png)
,解得:d=1-3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/20.png)
,此時,兩個三角形不相似;
⑤當P在y軸上時,AC是直角邊,P一定在B的左側,設P的坐標是(e,0).
則AP=1-e,當AC與DC是對應邊時,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/21.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/22.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/23.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/24.png)
,解得:e=-9,符合條件.
總之,符合條件的點P的坐標為:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193529177626577/SYS201311011935291776265025_DA/25.png)
.
點評:本題是相似三角形的判定與性質,待定系數(shù)法,勾股定理以及其逆定理的綜合應用.