【答案】
分析:(1)根據(jù)題意與圖象可得點(diǎn)C的坐標(biāo),根據(jù)圓的性質(zhì)可得點(diǎn)B的坐標(biāo),根據(jù)對(duì)稱軸方程與點(diǎn)B的坐標(biāo)即可求得函數(shù)的解析式;
(2)由拋物線的解析式可求得點(diǎn)A,E,B,C,D的坐標(biāo),判斷Rt△BOD∽R(shí)t△BCE,得∠CBE=∠OBD=β,因此sin(α-β)=sin(∠DBC-∠OBD)=sin∠OBC=

;
(3)顯然Rt△COA∽R(shí)t△BCE,此時(shí)點(diǎn)P
1(0,0),
過(guò)A作AP
2⊥AC交y正半軸于P
2,由Rt△CAP
2∽R(shí)t△BCE,得P
2(0,

),
過(guò)C作CP
3⊥AC交x正半軸于P
3,由Rt△P
3CA∽R(shí)t△BCE,得P
3(9,0),
故在坐標(biāo)軸上存在三個(gè)點(diǎn)P
1(0,0),P
2(0,
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),P
3(9,0),使得以P、A、C為頂點(diǎn)的三角形與△BCE相似.
解答:
解:(1)由題意可知C(0,-3),-
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=1,
∴拋物線的解析式為y=ax
2-2ax-3(a>0),
過(guò)M作MN⊥y軸于N,連接CM,則MN=1,CM=
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,
∴CN=2,于是m=-1.
同理可求得B(3,0),
∴a×3
2-2a×3-3=0,得a=1.
∴拋物線的解析式為y=x
2-2x-3.
(2)由(1)得A(-1,0),E(1,-4),B(3,0),C(0,-3).
∵M(jìn)到AB,CD的距離相等,OB=OC,
∴OA=OD,
∴點(diǎn)D的坐標(biāo)為(0,1),
∴在Rt△BCO中,BC=

=3

,
∴

,
在△BCE中,∵BC
2+CE
2=(3
2+3
2)+[(1-0)
2+(-4+3)
2]=20=(3-1)
2+(0+4)
2=BE
2
∴△BCE是Rt△

,
∴

,
即

,
∴Rt△BOD∽R(shí)t△BCE,得∠CBE=∠OBD=β,
因此sin(α-β)=sin(∠DBC-∠OBD)=sin∠OBC=

.
(3)顯然Rt△COA∽R(shí)t△BCE,此時(shí)點(diǎn)P
1(0,0).
過(guò)A作AP
2⊥AC交y正半軸于P
2,
由Rt△CAP
2∽R(shí)t△BCE,得P
2(0,

).
過(guò)C作CP
3⊥AC交x正半軸于P
3,由Rt△P
3CA∽R(shí)t△BCE,得P
3(9,0).
故在坐標(biāo)軸上存在三個(gè)點(diǎn)P
1(0,0),P
2(0,

),P
3(9,0),
使得以P、A、C為頂點(diǎn)的三角形與△BCE相似.
點(diǎn)評(píng):此題考查了二次函數(shù)與圓的知識(shí)的綜合應(yīng)用,要注意分析圖形,應(yīng)用相似三角形的性質(zhì)與判定,要注意數(shù)形結(jié)合思想的應(yīng)用.