【答案】
分析:(1)由題意拋物線y=ax
2+bx+c(a>0)的圖象經(jīng)過點(diǎn)B(12,0)和C(0,-6),對稱軸為x=2,根據(jù)待定系數(shù)法可以求得該拋物線的解析式;
(2)假設(shè)存在,設(shè)出時(shí)間t,則根據(jù)線段PQ被直線CD垂直平分,再由垂直平分線的性質(zhì)及勾股定理來求解t,看t是否存在;
(3)假設(shè)直線x=1上是存在點(diǎn)M,使△MPQ為等腰三角形,此時(shí)要分兩種情況討論:①當(dāng)PQ為等腰△MPQ的腰時(shí),且P為頂點(diǎn);②當(dāng)PQ為等腰△MPQ的腰時(shí),且Q為頂點(diǎn);然后再根據(jù)等腰三角形的性質(zhì)及直角三角形的勾股定理求出M點(diǎn)坐標(biāo).
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/images0.png)
解:(1)方法一:∵拋物線過C(0,-6)
∴c=-6,即y=ax
2+bx-6
由
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/0.png)
解得:a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/1.png)
,b=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/2.png)
∴該拋物線的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/3.png)
(3分)
方法二:∵A、B關(guān)于x=2對稱
∴A(-8,0)
設(shè)y=a(x+8)(x-12)
C在拋物線上
∴-6=a×8×(-12)
即a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/4.png)
∴該拋物線的解析式為:y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/5.png)
;(3分)
(2)存在,設(shè)直線CD垂直平分PQ,
在Rt△AOC中,AC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/6.png)
=10=AD,
∴點(diǎn)D在對稱軸上,連接DQ,顯然∠PDC=∠QDC (1分)
由已知∠PDC=∠ACD,
∴∠QDC=∠ACD,
∴DQ∥AC (1分)
∴DB=AB-AD=20-10=10,
∴DQ為△ABC的中位線,
∴DQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/7.png)
AC=5,(1分)
∴AP=AD-PD=AD-DQ=10-5=5,
∴t=5÷1=5(秒),
∴存在t=5(秒)時(shí),線段PQ被直線CD垂直平分(1分)
在Rt△BOC中,BC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/8.png)
,
而DQ為△ABC的中位線,
∴CQ=3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/9.png)
,
∴點(diǎn)Q的運(yùn)動(dòng)速度為每秒
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/10.png)
單位長度;(1分)
(3)存在,過點(diǎn)Q作QH⊥x軸于H,則QH=3,PH=9
在Rt△PQH中,PQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/11.png)
(1分)
①當(dāng)MP=MQ,即M為頂點(diǎn),
設(shè)直線CD的直線方程為:y=kx+b(k≠0),
則:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/12.png)
解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/13.png)
∴y=3x-6
當(dāng)x=1時(shí),y=-3,
∴M
1(1,-3)(1分)
②當(dāng)PQ為等腰△MPQ的腰時(shí),且P為頂點(diǎn).
設(shè)直線x=1上存在點(diǎn)M(1,y),
則OP=3,點(diǎn)M的橫坐標(biāo)為1,縱坐標(biāo)為y,根據(jù)勾股定理得PM
22=4
2+y
2,
又PQ
2=90,
則4
2+y
2=90,
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/14.png)
∴M
2(1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/15.png)
),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/16.png)
(1分)
③當(dāng)PQ為等腰△MPQ的腰時(shí),且Q為頂點(diǎn),
過點(diǎn)Q作QE⊥y軸于E,交直線x=1于F,則F(1,-3)
設(shè)直線x=1存在點(diǎn)M(1,y),由勾股定理得:
(y+3)
2+5
2=90即y=-3
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/17.png)
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/18.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/19.png)
(1分)
綜上所述:存在這樣的五點(diǎn):
M
1(1,-3),M
2(1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/20.png)
),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/21.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/22.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191854998801268/SYS201311011918549988012023_DA/23.png)
.
點(diǎn)評:此題是一道綜合題,難度較大,主要考查二次函數(shù)的性質(zhì),用待定系數(shù)法求函數(shù)的解析式,還考查等腰三角形的性質(zhì)及勾股定理,同時(shí)還讓學(xué)生探究存在性問題,對待問題要思考全面,學(xué)會分類討論的思想.