【答案】
分析:(1)利用拋物線對稱軸求出a的值,從而得解;
(2)根據(jù)拋物線解析式求出點(diǎn)A、B的坐標(biāo),然后利用待定系數(shù)法求一次函數(shù)解析式求出直線AB的解析式,再根據(jù)互相平行的直線的解析式的k值相等求出直線l的解析式,再根據(jù)軸對稱的性質(zhì)求出點(diǎn)A關(guān)于直線l的對稱點(diǎn)A′,連接A′B交直線l于點(diǎn)P,然后根據(jù)中點(diǎn)公式利用點(diǎn)A′、B的坐標(biāo)求出點(diǎn)P即可;
(3)根據(jù)直線l的解析式可得∠POB=45°,再求出OP、AB的長度,然后分①∠BAQ=∠POB=45°時(shí),②∠ABQ=∠POB=45°時(shí),根據(jù)相似三角形對應(yīng)邊成比例列式求出AQ的長度,從而得解.
解答:解:(1)∵對稱軸為直線x=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/0.png)
=4,
∴a=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/1.png)
,
∴拋物線解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/2.png)
x
2+2x;
(2)∵y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/3.png)
x
2+2x=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/4.png)
(x
2-8x+16)+4=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/5.png)
(x-4)
2+4,
∴頂點(diǎn)坐標(biāo)為A(4,4),
令y=0,則-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/6.png)
x
2+2x=0,
解得x
1=0,x
2=8,
∴點(diǎn)B的坐標(biāo)為(8,0),
設(shè)直線AB的解析式為y=kx+b,
則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/7.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/8.png)
,
所以,直線AB的解析式為y=-x+8,
∵直線l為直線AB平移至經(jīng)過原點(diǎn)的直線,
∴直線l的解析式為y=-x,
如圖,取點(diǎn)A關(guān)于直線l的對稱點(diǎn)A′,連接A′B交直線l于點(diǎn)P,則△PAB的周長最小,
此時(shí),點(diǎn)A(-4,-4),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/images9.png)
點(diǎn)P為線段A′B的中點(diǎn),
∵
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/9.png)
=2,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/10.png)
=-2,
∴點(diǎn)P的坐標(biāo)為(2,-2);
(3)∵直線AB的解析式為y=-x+8,
∴直線AB與x軸、對稱軸的夾角的銳角為45°,
又∵l∥AB,
∴∠POB=45°,
根據(jù)勾股定理,AB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/11.png)
=4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/12.png)
,
PO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/13.png)
=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/14.png)
,
①∠BAQ=∠POB=45°時(shí),∵△POB∽△BAQ,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/15.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/16.png)
,
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/17.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/18.png)
,
解得AQ=16,
∴Q的橫坐標(biāo)為16+4=20,縱坐標(biāo)為4,
∴點(diǎn)Q的坐標(biāo)為(20,4);
②∠ABQ=∠POB=45°時(shí),∵△POB∽△ABQ,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/19.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/20.png)
,
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/21.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202408604480957/SYS201311032024086044809022_DA/22.png)
,
解得BQ=16,
∴點(diǎn)Q的坐標(biāo)為(8,16),
綜上所述,存在點(diǎn)Q(20,4)或(8,16)使以A,B,Q為頂點(diǎn)的三角形與△POB相似.
點(diǎn)評:本題是二次函數(shù)綜合題型,主要利用了拋物線的對稱軸公式,拋物線的對應(yīng)點(diǎn)坐標(biāo),與x軸的交點(diǎn)坐標(biāo),利用軸對稱確定最短路線問題,相似三角形對應(yīng)邊成比例的性質(zhì),(3)根據(jù)直線的解析式確定出45°是解題的關(guān)鍵,要注意分情況討論求解.