【答案】
分析:(1)已知了A、B兩點(diǎn)的坐標(biāo)即可得出OA、OB的長,在直角三角形ACB中由于OC⊥AB,因此可用射影定理求出OC的長,即可得出C點(diǎn)的坐標(biāo).然后用待定系數(shù)法即可求出拋物線的解析式;
(2)本題的關(guān)鍵是得出D點(diǎn)的坐標(biāo),CD平分∠BCE,如果連接O′D,那么根據(jù)圓周角定理即可得出∠DO′B=2∠BCD=∠BCE=90°由此可得出D的坐標(biāo)為(4,-5).根據(jù)B、D兩點(diǎn)的坐標(biāo)即可用待定系數(shù)法求出直線BD的解析式;
(3)本題要分兩種情況進(jìn)行討論:
①過D作DP∥BC,交D點(diǎn)右側(cè)的拋物線于P,此時(shí)∠PDB=∠CBD,可先用待定系數(shù)法求出直線BC的解析式,然后根據(jù)BC與DP平行,那么直線DP的斜率與直線BC的斜率相同,因此可根據(jù)D的坐標(biāo)求出DP的解析式,然后聯(lián)立直線DP的解析式和拋物線的解析式即可求出交點(diǎn)坐標(biāo),然后將不合題意的舍去即可得出符合條件的P點(diǎn).
②同①的思路類似,先作與∠CBD相等的角:在O′B上取一點(diǎn)N,使BN=BM.可通過證△NBD≌△MDB,得出∠NDB=∠CBD,然后同①的方法一樣,先求直線DN的解析式,進(jìn)而可求出其與拋物線的交點(diǎn)即P點(diǎn)的坐標(biāo).綜上所述可求出符合條件的P點(diǎn)的值.
解答:解:(1)∵以AB為直徑作⊙O′,交y軸的負(fù)半軸于點(diǎn)C,
∴∠OCA+∠OCB=90°,
又∵∠OCB+∠OBC=90°,
∴∠OCA=∠OBC,
又∵∠AOC=∠COB=90°,
∴△AOC∽△COB,
∴
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.
又∵A(-1,0),B(9,0),
∴
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,
解得OC=3(負(fù)值舍去).
∴C(0,-3),
故設(shè)拋物線解析式為y=a(x+1)(x-9),
∴-3=a(0+1)(0-9),解得a=
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,
∴二次函數(shù)的解析式為y=
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(x+1)(x-9),
即y=
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x
2-
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x-3.
(2)∵AB為O′的直徑,且A(-1,0),B(9,0),
∴OO′=4,O′(4,0),
∵點(diǎn)E是AC延長線上一點(diǎn),∠BCE的平分線CD交⊙O′于點(diǎn)D,
∴∠BCD=
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∠BCE=
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×90°=45°,
連接O′D交BC于點(diǎn)M,
則∠BO′D=2∠BCD=2×45°=90°,OO′=4,O′D=
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AB=5.
∴O′D⊥x軸
∴D(4,-5).
∴設(shè)直線BD的解析式為y=kx+b,
∴
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,
解得
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∴直線BD的解析式為y=x-9.
∵C(0,-3),
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設(shè)直線BC的解析式為:y=ax+b,
∴
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,
解得:
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,
∴直線BC的解析式為:y=
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x-3.
(3)假設(shè)在拋物線上存在點(diǎn)P,使得∠PDB=∠CBD,
解法一:設(shè)射線DP交⊙O′于點(diǎn)Q,則
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=
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.
分兩種情況(如圖所示):
①∵O′(4,0),D(4,-5),B(9,0),C(0,-3).
∴把點(diǎn)C、D繞點(diǎn)O′逆時(shí)針旋轉(zhuǎn)90°,使點(diǎn)D與點(diǎn)B重合,則點(diǎn)C與點(diǎn)Q
1重合,
因此,點(diǎn)Q
1(7,-4)符合
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=
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,
∵D(4,-5),Q
1(7,-4),
∴用待定系數(shù)法可求出直線DQ
1解析式為y=
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x-
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.
解方程組
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得
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∴點(diǎn)P
1坐標(biāo)為(
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,

),坐標(biāo)為(
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,
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)不符合題意,舍去.
②∵Q
1(7,-4),
∴點(diǎn)Q
1關(guān)于x軸對(duì)稱的點(diǎn)的坐標(biāo)為Q
2(7,4)也符合
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=
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.
∵D(4,-5),Q
2(7,4).
∴用待定系數(shù)法可求出直線DQ
2解析式為y=3x-17.
解方程組
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得
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,
即
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∴點(diǎn)P
2坐標(biāo)為(14,25),坐標(biāo)為(3,-8)不符合題意,舍去.
∴符合條件的點(diǎn)P有兩個(gè):P
1(
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,
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),P
2(14,25).
解法二:分兩種情況(如圖所示):
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①當(dāng)DP
1∥CB時(shí),能使∠PDB=∠CBD.
∵B(9,0),C(0,-3).
∴用待定系數(shù)法可求出直線BC解析式為y=
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x-3.
又∵DP
1∥CB,
∴設(shè)直線DP
1的解析式為y=
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x+n.
把D(4,-5)代入可求n=-
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,
∴直線DP
1解析式為y=
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x-
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.
解方程組
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得
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∴點(diǎn)P
1坐標(biāo)為(
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,
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)或(
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,
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)(不符合題意舍去).
②在線段O′B上取一點(diǎn)N,使BN=DM時(shí),得△NBD≌△MDB(SAS),
∴∠NDB=∠CBD.
由①知,直線BC解析式為y=
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x-3.
取x=4,得y=-
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,
∴M(4,-
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),
∴O′N=O′M=
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,
∴N(
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,0),
又∵D(4,-5),
∴直線DN解析式為y=3x-17.
解方程組
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得
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,
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∴點(diǎn)P
2坐標(biāo)為(14,25),坐標(biāo)為(3,-8)不符合題意,舍去.
∴符合條件的點(diǎn)P有兩個(gè):P
1(
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,
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),P
2(14,25).
解法三:分兩種情況(如圖所示):
①求點(diǎn)P
1坐標(biāo)同解法二.

②過C點(diǎn)作BD的平行線,交圓O′于G,
此時(shí),∠GDB=∠GCB=∠CBD.
由(2)題知直線BD的解析式為y=x-9,
又∵C(0,-3)
∴可求得CG的解析式為y=x-3,
設(shè)G(m,m-3),作GH⊥x軸交于x軸與H,
連接O′G,在Rt△O′GH中,利用勾股定理可得,m=7,
由D(4,-5)與G(7,4)可得,
DG的解析式為y=3x-17,
解方程組
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得
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,
即
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∴點(diǎn)P
2坐標(biāo)為(14,25),坐標(biāo)為(3,-8)不符合題意舍去.
∴符合條件的點(diǎn)P有兩個(gè):P
1(
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,
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),P
2(14,25).
點(diǎn)評(píng):本題著重考查了待定系數(shù)法求二次函數(shù)解析式、三角形相似及全等、探究角相等的構(gòu)成情況等知識(shí)點(diǎn),綜合性強(qiáng),考查學(xué)生分類討論,數(shù)形結(jié)合的數(shù)學(xué)思想方法.