【答案】
分析:(1)已知拋物線的頂點(diǎn)為A(2,1),設(shè)拋物線頂點(diǎn)式,把點(diǎn)O(0,0)代入即可求解析式;
(2)依題意得CD∥OB,CD=OB=4,又對(duì)稱軸x=2,故D點(diǎn)橫坐標(biāo)x=6,代入拋物線解析式可求D點(diǎn)縱坐標(biāo),根據(jù)對(duì)稱軸可求滿足條件的點(diǎn)D′;
(3)根據(jù)拋物線對(duì)稱軸可知AO=AB,△AOB為等腰三角形,要使得△OBP與△OAB相似,則∠POB=∠BOA,A與A′對(duì)稱,可求直線OP的解析式,與拋物線解析式聯(lián)立可求P點(diǎn)坐標(biāo),檢驗(yàn)BP與OB是否相等.
解答:解:(1)由題意可設(shè)拋物線的解析式為
y=a(x-2)
2+1
∵拋物線過原點(diǎn),
∴0=a(0-2)
2+1,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348026_DA/0.png)
.
拋物線的解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348026_DA/1.png)
(x-2)
2+1,
即y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348026_DA/2.png)
x
2+x
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348026_DA/images3.png)
(2)如圖1,當(dāng)四邊形OCDB是平行四邊形時(shí),CD=OB,
由0=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348026_DA/3.png)
(x-2)
2+1得x
1=0,x
2=4,
∴B(4,0),OB=4.
由于對(duì)稱軸x=2
∴D點(diǎn)的橫坐標(biāo)為6.
將x=6代入y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348026_DA/4.png)
(x-2)
2+1,得y=-3,
∴D(6,-3);
根據(jù)拋物線的對(duì)稱性可知,
在對(duì)稱軸的左側(cè)拋物線上存在點(diǎn)D,使得四邊形ODCB是平行四邊形,此時(shí)D點(diǎn)的坐標(biāo)為(-2,-3),
當(dāng)四邊形OCBD是平行四邊形時(shí),D點(diǎn)即為A點(diǎn),此時(shí)D點(diǎn)的坐標(biāo)為(2,1)
(3)不存在.
如圖2,由拋物線的對(duì)稱性可知:AO=AB,∠AOB=∠ABO.
若△BOP與△AOB相似,必須有∠POB=∠BOA=∠BPO
設(shè)OP交拋物線的對(duì)稱軸于A′點(diǎn),顯然A′(2,-1)
∴直線OP的解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348026_DA/5.png)
x
由-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348026_DA/6.png)
x=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348026_DA/7.png)
x
2+x,得x
1=0,x
2=6.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348026_DA/images9.png)
∴P(6,-3)
過P作PE⊥x軸,在Rt△BEP中,BE=2,PE=3,
∴PB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101618660834891/SYS201311031016186608348026_DA/8.png)
≠4.
∴PB≠OB,
∴∠BOP≠∠BPO,
∴△PBO與△BAO不相似,
同理可說明在對(duì)稱軸左邊的拋物線上也不存在符合條件的P點(diǎn).
所以在該拋物線上不存在點(diǎn)P,使得△BOP與△AOB相似.
點(diǎn)評(píng):本題考查了二次函數(shù)解析式的求法,利用拋物線的性質(zhì)尋找平行四邊形,相似三角形等問題,需要根據(jù)拋物線的對(duì)稱性,形數(shù)結(jié)合,解答問題.