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解:(1)連接AC,由直線BC為圓A的切線,得到CA⊥CB,
又∵⊙A的半徑為3,
∴AC=3,
又∵A點(diǎn)的坐標(biāo)為(2,0),即OA=2,
在Rt△AOC中,根據(jù)勾股定理得:OC=
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=
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,
∴點(diǎn)C坐標(biāo)為(0,
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),
又∠OCB+∠OCA=90°,∠OCA+∠OAC=90°,
∴∠OCB=∠OAC,又∠COB=∠AOC=90°,
∴△BOC∽△COA,
∴
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=
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,又OC=
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,OA=2,
∴BO=
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,即B(-
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,0),
設(shè)直線BC的方程為y=kx+b,
把B和C的坐標(biāo)代入得:
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,
解得:k=
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,b=
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,
則直線BC的方程為y=
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x+
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;
(2)拋物線y=ax
2+bx+c經(jīng)過(guò)B、A兩點(diǎn),且頂點(diǎn)在直線BC上,
∵A(2,0),B(-
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,0),
∴
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=-
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,
∴對(duì)稱(chēng)軸為直線x=-
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,即頂點(diǎn)橫坐標(biāo)為-
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,
把x=-
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代入y=
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x+
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得:y=
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,
則此拋物線的頂點(diǎn)的坐標(biāo)為(-
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,
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);
(3)x軸上存在一點(diǎn)P,使△PCE和△CBE相似,理由如下:
∵AE=3,OA=2,
∴OE=1,
在Rt△OCE中,根據(jù)勾股定理得:CE=
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=
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,
∵OB=
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,OE=1,
∴BE=1.5,
假設(shè)存在這樣的點(diǎn)P,
當(dāng)點(diǎn)P在點(diǎn)B左側(cè)時(shí),如圖所示:
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若△BCE∽△CPE,則有
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,
即
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=
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,
解得:PE=4,
則點(diǎn)P的坐標(biāo)為(-5,0);
當(dāng)點(diǎn)P在點(diǎn)B右側(cè)時(shí),要使△CBE∽△PCE,則有∠BEC=∠CEP,
∴∠BEC=∠CEP=90°,與題設(shè)矛盾,
∴不存在這樣的P滿(mǎn)足題意,
綜上,滿(mǎn)足題意的P點(diǎn)有1個(gè),P的坐標(biāo)為(-5,0).
分析:(1)連接AC,由BC為圓A的切線,根據(jù)切線的性質(zhì)得到CA與CB垂直,同時(shí)由半徑為3,得到AC為3,由A的坐標(biāo)得到OA的長(zhǎng),在直角三角形OAC中,利用勾股定理求出OC的長(zhǎng),確定出C的坐標(biāo),又CO與AB垂直,得到一對(duì)直角相等,再利用同角的余角相等,得到一對(duì)角相等,根據(jù)兩對(duì)對(duì)應(yīng)角相等的兩三角形相似可得三角形BOC與三角形COA相似,由相似得線段成比例,將OC,OA的長(zhǎng)代入求出OB的長(zhǎng),確定出B的坐標(biāo),設(shè)出直線BC的方程為y=kx+b,將B和C的坐標(biāo)代入,得出關(guān)于k與b的方程組,求出方程組的解集得到k與b的值,確定出直線BC的方程;
(2)拋物線y=ax
2+bx+c經(jīng)過(guò)B、A兩點(diǎn),且頂點(diǎn)在直線BC上,由A和B的坐標(biāo)求出線段AB的中點(diǎn)坐標(biāo),即為拋物線頂點(diǎn)的橫坐標(biāo),將求出的橫坐標(biāo)代入直線BC的解析式中求出對(duì)應(yīng)y的值,即為頂點(diǎn)的縱坐標(biāo),進(jìn)而確定出頂點(diǎn)的坐標(biāo);
(3)存在,理由為:假設(shè)在x軸上存在一點(diǎn)P,使△PCE和△CBE相似,可能有兩種情況,當(dāng)P在B的左側(cè)時(shí),根據(jù)題意畫(huà)出圖形,由AE為圓的半徑,OA,利用AE-OA求出OE的長(zhǎng),在直角三角形COE中,由OC及OE的長(zhǎng),利用勾股定理求出EC的長(zhǎng),再由OB-OE求出BE的長(zhǎng),根據(jù)△PCE和△CBE相似得出比例式,將求出的BE及CE代入,求出PE的長(zhǎng),根據(jù)PE+OE=OP,求出OP的長(zhǎng),可得出此時(shí)P的坐標(biāo);當(dāng)P在B的右側(cè)時(shí),由△PCE和△CBE相似,得到∠BEC=∠ECP,根據(jù)內(nèi)錯(cuò)角相等兩直線平行得到CP與x軸平行,即CP與x軸沒(méi)有交點(diǎn),此時(shí)P不存在,綜上,得到滿(mǎn)足題意的P的坐標(biāo).
點(diǎn)評(píng):此題屬于二次函數(shù)的綜合題,涉及的知識(shí)有:切線的性質(zhì),相似三角形的判定與性質(zhì),二次函數(shù)的性質(zhì),勾股定理,利用待定系數(shù)法求一次函數(shù)的解析式,利用了轉(zhuǎn)化,分類(lèi)討論及數(shù)形結(jié)合的思想,是一道綜合性較強(qiáng)的題.本題比較難,鍛煉了學(xué)生綜合分析問(wèn)題,解決問(wèn)題的能力.