【答案】
分析:(1)由題意拋物線y=ax
2+bx+c(a>0)的圖象經(jīng)過(guò)點(diǎn)B(12,0)和C(0,-6),對(duì)稱(chēng)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(lái)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)方法一:∵拋物線過(guò)C(0,-6)
∴c=-6,即y=ax
2+bx-6
由
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解得:a=
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,b=-
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∴該拋物線的解析式為y=
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(3分)
方法二:∵A、B關(guān)于x=2對(duì)稱(chēng)
∴A(-8,0)
設(shè)y=a(x+8)(x-12)
C在拋物線上
∴-6=a×8×(-12)
即a=

∴該拋物線的解析式為:y=

;(3分)
(2)存在,設(shè)直線CD垂直平分PQ,
在Rt△AOC中,AC=
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=10=AD,
∴點(diǎn)D在對(duì)稱(chēng)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=

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=

,
而DQ為△ABC的中位線,
∴CQ=3

,
∴點(diǎn)Q的運(yùn)動(dòng)速度為每秒

單位長(zhǎng)度;(1分)
(3)存在,過(guò)點(diǎn)Q作QH⊥x軸于H,則QH=3,PH=9
在Rt△PQH中,PQ=

(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:
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解得:
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∴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
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∴M
2(1,
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),
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(1分)
③當(dāng)PQ為等腰△MPQ的腰時(shí),且Q為頂點(diǎn),
過(guò)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

∴
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
(1分)
綜上所述:存在這樣的五點(diǎn):
M
1(1,-3),M
2(1,
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),
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.
點(diǎn)評(píng):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(wèn)題,對(duì)待問(wèn)題要思考全面,學(xué)會(huì)分類(lèi)討論的思想.