【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(14,0)和C(0,-8),對稱軸為x=4,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:解:(1)∵拋物線過C(0,-8),
∴c=-8,即y=ax
2+bx-8,
由函數(shù)經(jīng)過點(diǎn)(14,0)及對稱軸為x=4可得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/0.png)
,
解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/1.png)
,
∴該拋物線的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/2.png)
x
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/3.png)
x-8.
(2)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/images4.png)
存在直線CD垂直平分PQ.
由函數(shù)解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/4.png)
x
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/5.png)
x-8,可求出點(diǎn)A坐標(biāo)為(-6,0),
在Rt△AOC中,AC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/6.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/7.png)
=10=AD,
故可得OD=AD-OA=4,點(diǎn)D在函數(shù)的對稱軸上,
∵線CD垂直平分PQ,
∴∠PDC=∠QDC,PD=DQ,
由AD=AC可得,∠PDC=∠ACD,
∴∠QDC=∠ACD,
∴DQ∥AC,
又∵DB=AB-AD=20-10=10=AD,
∴點(diǎn)D是AB中點(diǎn),
∴DQ為△ABC的中位線,
∴DQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/8.png)
AC=5,
∴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.
在Rt△BOC中,BC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/9.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/10.png)
=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/11.png)
,
而DQ為△ABC的中位線,Q是BC中點(diǎn),
∴CQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/12.png)
,
∴點(diǎn)Q的運(yùn)動(dòng)速度為每秒
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/13.png)
單位長度;
(3)存在,過點(diǎn)Q作QH⊥x軸于H,則QH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/14.png)
OC=4,PH=OP+OH=1+7=8,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/images16.png)
在Rt△PQH中,PQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/15.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/16.png)
=4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/17.png)
,
①當(dāng)MP=MQ,即M為頂點(diǎn),則此時(shí)CD與PQ的交點(diǎn)即是M點(diǎn)(上面已經(jīng)證明CD垂直平分PQ),
設(shè)直線CD的直線方程為:y=kx+b(k≠0),
因?yàn)辄c(diǎn)C(0,-8),點(diǎn)D(4,0),
所以可得直線CD的解析式為:y=2x-8,
當(dāng)x=1時(shí),y=-6,
∴M
1(1,-6);
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),因?yàn)辄c(diǎn)P坐標(biāo)為(-1,0),
從而可得PM
2=2
2+y
2,
又PQ
2=80,
則2
2+y
2=80,
即y=±
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/18.png)
,
∴M
2(1,2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/19.png)
),M
3(1,-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/20.png)
);
③當(dāng)PQ為等腰△MPQ的腰時(shí),且Q為頂點(diǎn),點(diǎn)Q坐標(biāo)為(7,-4),
設(shè)直線x=1存在點(diǎn)M(1,y),
則QM
2=6
2+(y+4)
2=80,
解得:y=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/21.png)
-4或-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/22.png)
-4;
∴M
4(1,-4+2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/23.png)
),M
5(1,-4-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/24.png)
);
綜上所述:存在這樣的五點(diǎn):
M
1(1,-6),M
2(1,2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/25.png)
),M
3(1,-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/26.png)
)M
4(1,-4+2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/27.png)
),M
5(1,-4-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103202255044322281/SYS201311032022550443222024_DA/28.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),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(huì)分類討論的思想.