【答案】
分析:(1)根據(jù)對(duì)稱(chēng)軸公式,A、C兩點(diǎn)坐標(biāo),列方程組,求拋物線解析式;
(2)①只需要AP∥BC即可滿(mǎn)足題意,先求直線BC解析式,根據(jù)平行線的解析式一次項(xiàng)系數(shù)相等,設(shè)直線AP的解析式,將A點(diǎn)坐標(biāo)代入可求直線AP的解析式,將拋物線與直線AP解析式聯(lián)立,即可求P點(diǎn)坐標(biāo),再根據(jù)平移法求滿(mǎn)足條件的另外兩個(gè)P點(diǎn)坐標(biāo);
②延長(zhǎng)CP交x軸于點(diǎn)Q,根據(jù)拋物線解析式可知△OBC為等腰直角三角形,利用角的關(guān)系證明∠OCA=∠OQC,可證Rt△AOC∽R(shí)t△COQ,利用相似比求解.
解答:解:(1)由題意,得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160210695281547/SYS201310221602106952815023_DA/0.png)
,解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160210695281547/SYS201310221602106952815023_DA/1.png)
∴拋物線的解析式為y=-x
2+4x-3;
(2)①令-x
2+4x-3=0,解得x
1=1,x
2=3,∴B(3,0),
當(dāng)點(diǎn)P在x軸上方時(shí),如圖1,
過(guò)點(diǎn)A作直線BC的平行線交拋物線于點(diǎn)P,
易求直線BC的解析式為y=x-3,
∴設(shè)直線AP的解析式為y=x+n,
∵直線AP過(guò)點(diǎn)A(1,0),代入求得n=-1.
∴直線AP的解析式為y=x-1
解方程組
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160210695281547/SYS201310221602106952815023_DA/2.png)
,得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160210695281547/SYS201310221602106952815023_DA/3.png)
,
∴點(diǎn)P
1(2,1)
當(dāng)點(diǎn)P在x軸下方時(shí),如圖1:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160210695281547/SYS201310221602106952815023_DA/images4.png)
設(shè)直線AP
1交y軸于點(diǎn)E(0,-1),
把直線BC向下平移2個(gè)單位,交拋物線于點(diǎn)P
2,P
3,
得直線P
2P
3的解析式為y=x-5,
解方程組
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160210695281547/SYS201310221602106952815023_DA/4.png)
,
得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160210695281547/SYS201310221602106952815023_DA/5.png)
,
∴P
2(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160210695281547/SYS201310221602106952815023_DA/6.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160210695281547/SYS201310221602106952815023_DA/7.png)
),P
3(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160210695281547/SYS201310221602106952815023_DA/8.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160210695281547/SYS201310221602106952815023_DA/9.png)
),
綜上所述,點(diǎn)P的坐標(biāo)為:P
1(2,1),P
2(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160210695281547/SYS201310221602106952815023_DA/10.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160210695281547/SYS201310221602106952815023_DA/11.png)
),P
3(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160210695281547/SYS201310221602106952815023_DA/12.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160210695281547/SYS201310221602106952815023_DA/13.png)
),
②∵B(3,0),C(0,-3)
∴OB=OC,∴∠OCB=∠OBC=45°
設(shè)直線CP的解析式為y=kx-3
如圖2,延長(zhǎng)CP交x軸于點(diǎn)Q,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160210695281547/SYS201310221602106952815023_DA/images15.png)
設(shè)∠OCA=α,則∠ACB=45°-α,
∵∠PCB=∠BCA,∴∠PCB=45°-α,
∴∠OQC=∠OBC-∠PCB=45°-(45°-α)=α,
∴∠OCA=∠OQC
又∵∠AOC=∠COQ=90°
∴Rt△AOC∽R(shí)t△COQ
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160210695281547/SYS201310221602106952815023_DA/14.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160210695281547/SYS201310221602106952815023_DA/15.png)
,
∴OQ=9,∴Q(9,0)
∵直線CP過(guò)點(diǎn)Q(9,0),∴9k-3=0
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160210695281547/SYS201310221602106952815023_DA/16.png)
∴直線CP的解析式為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022160210695281547/SYS201310221602106952815023_DA/17.png)
.
其它方法略.
點(diǎn)評(píng):本題考查了二次函數(shù)的綜合運(yùn)用.關(guān)鍵是由已知條件求拋物線解析式,根據(jù)拋物線與x軸,y軸的交點(diǎn),判斷三角形的特殊性,利用平移,相似的知識(shí)解題.