【答案】
分析:(1)欲求拋物線的解析式,需求出m、n的值,根據(jù)拋物線的解析式,易得頂點(diǎn)A的坐標(biāo),然后將x=1代入拋物線的解析式中,可得點(diǎn)C的坐標(biāo),即可根據(jù)AC的長(zhǎng)利用勾股定理得到第一個(gè)關(guān)于m、n的等量關(guān)系式;由于拋物線的頂點(diǎn)在x軸上,即拋物線與x軸只有一個(gè)交點(diǎn),即根的判別式△=0,聯(lián)立兩個(gè)關(guān)于m、n的式子即可求出m、n的值,從而得到該拋物線的解析式;
(2)根據(jù)(1)的拋物線解析式可求得點(diǎn)B的坐標(biāo),即可得到OB的長(zhǎng);過(guò)O作OM⊥BD于M,根據(jù)題意可知OM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/0.png)
,進(jìn)而可利用勾股定理求得BM的長(zhǎng);在△EOF中,OM⊥EF,易證得△OBM∽△FOM,根據(jù)相似三角形所得比例線段即可求得OF的長(zhǎng),也就得到了F點(diǎn)的坐標(biāo),進(jìn)而可利用待定系數(shù)法求得直線BD的解析式,聯(lián)立拋物線的解析式即可求出點(diǎn)D的坐標(biāo).
(3)存在.利用△ABF∽△AOB、△ABP
2∽△BOA、△ABP
3∽△BOA、△ABP
4∽△AOB可分別確定P
1、P
2、P
3、P
4的坐標(biāo).
解答:解:(1)過(guò)點(diǎn)C作CE⊥x軸于點(diǎn)E,如圖,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/images1.png)
∵拋物線上一點(diǎn)C的橫坐標(biāo)為1,
∴C(1,n-2m+2),
其中n-2m+2>0,OE=1,CE=n-2m+2;
∵拋物線的頂點(diǎn)A在x軸負(fù)半軸上,
∴A(m,0),△=4m
2-4(n+1)=0,得n=m
2-1①,
其中m<0,OA=-m,AE=OE+OA=1-m,
在Rt△ACE中,AC=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/1.png)
,
∵AE
2+CE
2=AC
2,
∴(1-m)
2+(n-2m+2)
2=(3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/2.png)
)
2②,
把①代入②得[(m-1)
2]
2+(m-1)
2-90=0,
∴[(m-1)
2+10][(m-1)
2-9]=0,
∴(m-1)
2-9=0
∴m
1=4,m
2=-2,
∵m<0,
∴m=-2.
把m=-2代入①,得n=4-1=3,
∴拋物線的關(guān)系式為y=x
2+4x+4;
(2)設(shè)直線DB交x軸正半軸于點(diǎn)F,過(guò)點(diǎn)O作OM⊥DB于點(diǎn)M,如圖,
∵點(diǎn)O到直線DB的距離為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/3.png)
,
∴OM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/4.png)
,
而B(niǎo)點(diǎn)坐標(biāo)為(0,4),
∴OB=4,
∴BM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/5.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/6.png)
;
∵OB⊥OF,OM⊥BF,
∴△OBM∽△FOM,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/7.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/8.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/9.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/10.png)
,
∴OF=8,
∴F點(diǎn)坐標(biāo)為(8,0),
設(shè)直線DB的解析式為y=kx+b,
把F(8,0)、B(0,4)代入得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/11.png)
,解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/12.png)
,
∴直線DB的解析式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/13.png)
x+4,
解方程組
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/14.png)
得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/15.png)
或
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/16.png)
,
∴D點(diǎn)坐標(biāo)為(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/17.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/18.png)
);
(3)存在.理由如下:
∵OB=4,OF=8,
∴BF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/19.png)
=4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/20.png)
,
∵y=(x+2)
2,
∴A點(diǎn)坐標(biāo)為(-2,0),
∴OA=2,
而OB=4,
∴AB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/21.png)
=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/22.png)
∴OA:OB=OB:OF,
∴△OAB∽△OBF,
∴∠AOB=∠OFB,
∴∠ABF=∠ABO+∠OBF=∠OFB+∠OBF=90°,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/images24.png)
∴△ABF∽△AOB,
此時(shí)P
1在F點(diǎn)位置,符號(hào)要求,P
1點(diǎn)的坐標(biāo)為(8,0);
當(dāng)△ABP
2∽△BOA時(shí),
則BP
2:OA=AB:BO,即BP
2:2=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/23.png)
:4,
∴BP
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/24.png)
,
過(guò)P
2作P
2H⊥x軸于H,如圖,
∴OH:OF=BP
2:BF,即OH:8=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/25.png)
:4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/26.png)
,
∴OH=2,
把x=2代入y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/27.png)
x+4得y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/28.png)
×2+4=2,
∴P
2的坐標(biāo)為(2,2);
當(dāng)△ABP
3∽△BOA時(shí),同樣得到BP
3=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/29.png)
,
∴P
3A⊥OA,
∴P
3的橫坐標(biāo)為-2,
把x=-2代入y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/30.png)
x+4得y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/31.png)
×(-2)+4=5,
∴P
3的坐標(biāo)為(-2,6);
當(dāng)△ABP
4∽△AOB時(shí),
則BP
4:OB=AB:AO,即BP
4:4=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/32.png)
:2,
∴BP
4=4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/33.png)
,
過(guò)P
4作P
4Q⊥y軸于Q,如圖,
易證得△P
4QB≌△FOB,
∴P
4Q=8,
把x=-8代入y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/34.png)
x+4得y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101193542900520873/SYS201311011935429005208022_DA/35.png)
×(-8)+4=8,
∴P
4的坐標(biāo)為(-8,8),
∴滿(mǎn)足條件的P點(diǎn)坐標(biāo)為(-8,8)、(-2,5)、(2,2)、(8,0).
點(diǎn)評(píng):此題是二次函數(shù)的綜合題,涉及到勾股定理、根的判別式、二次函數(shù)解析式的確定、相似三角形的判定和性質(zhì)以及函數(shù)圖象交點(diǎn)坐標(biāo)的求法等重要知識(shí),綜合性強(qiáng),難度較大.