【答案】
分析:(1)本題可通過(guò)全等三角形來(lái)證簡(jiǎn)單的線段相等,三角形ABF和ADO中,根據(jù)圓周角定理可得出∠ABF=∠ADO,已知了一組直角和AB=AD,因此兩三角形全等,即可得出BF=OD的結(jié)論;
(2)如果G是三角形BDO的外心,根據(jù)三角形外心定義可知BE必垂直平分OD,因此三角形BOD是等腰三角形.在等腰直角三角形ABD中,BD=BO=2
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,AB=OB-OA=2
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+m,因此可根據(jù)AB、BD的比例關(guān)系求出m的值,即可得出OA的長(zhǎng),而在(1)得出的全等三角形中,可得出OA=FG,據(jù)此可求出F點(diǎn)坐標(biāo).已知了B、F、O三點(diǎn)坐標(biāo),可用待定系數(shù)法求出拋物線的解析式;
(3)當(dāng)直線BE與y軸相交于G,向上平移直線BE使平移后的直線經(jīng)過(guò)原點(diǎn)O,由圖象知,在平移前直線BE與新圖象有1個(gè)公共點(diǎn),平移到經(jīng)過(guò)點(diǎn)O時(shí)與新圖象有3個(gè)公共點(diǎn),并且0<t<OG,利用已知條件求出OG的長(zhǎng)即可求出t的取值范圍;當(dāng)直線BE向上平移至于拋物線相切后再向上平移時(shí),直線BE與圖象的交點(diǎn)又變?yōu)閮蓚€(gè),設(shè)相切時(shí)直線BE的解析式為
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,求出方程組的解,進(jìn)而求出t的取值范圍.
解答:解:(1)∵四邊形ABCD是正方形,
∴AB=AD,∠BAF=∠DAO=90°
在△ABF和△ADO中
∵∠ABF=∠ADO,AB=AD,∠BAF=∠DAO
∴△ABF≌△ADO
∴BF=DO;
(2)∵A(m,0),B(
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),
∴AO=m,BO=
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,AB=
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m,
∵弧AE=弧DE,
∴∠EBO=∠EBD,
∵∠DAB=90°,
∴BD為直徑∴∠BEO=∠BED=90°,
又∵BE=BE,
∴△BEO≌△BED,
∴BD=BO=
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,
在Rt△BCD中BD=
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AB,
∴
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=
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,
∴m=
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,
∵△ABF≌△ADO,
∴AF=AO=m=
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,
∴F點(diǎn)的坐標(biāo)為
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,
∵拋物線l經(jīng)過(guò)O(0,0),B(
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),
設(shè)l的解析式為
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,
將F
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代入得:
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,
∴拋物線l的解析式為
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;
(3)①如圖,設(shè)直線BE與y軸相交于G,向上平移直線BE使平移后的直線經(jīng)過(guò)原點(diǎn)O,由圖象知,在平移前直線BE與新圖象有1個(gè)公共點(diǎn),平移到經(jīng)過(guò)點(diǎn)O時(shí)與新圖象有3個(gè)公共點(diǎn).∴0<t<OG
設(shè)直線BE的解析式為y=kx+m,將B(
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),F(xiàn)
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代入易求出:
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,
當(dāng)x=0時(shí),
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,
∴
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,
此時(shí)t的取值范圍是:
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.
②如圖,當(dāng)直線BE向上平移至于拋物線相切后再向上平移時(shí),直線BE與圖象的交點(diǎn)又變?yōu)閮蓚€(gè),設(shè)相切時(shí)直線BE的解析式為
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,則方程組
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有一個(gè)解,
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于是方程
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有兩個(gè)相等的實(shí)數(shù)根,求出
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,
此時(shí)直線BE的解析式為
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,
直線BE與y軸的交點(diǎn)為(0,
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)
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,
∴此時(shí)t的取值范圍是:
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.
綜上所述:t的取值范圍為:
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或
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.
點(diǎn)評(píng):本題考查了二次函數(shù)的性質(zhì),二次函數(shù)和圓的交點(diǎn)問(wèn)題,以及正方形的性質(zhì)和全等三角形的判定和全等三角形的性質(zhì),本題有一定的難度,綜合性也比較強(qiáng),有一定的新意,第3小問(wèn)有些難度,有一定的能力要求,解這種題時(shí)需冷靜地分析題意,找到切入點(diǎn)不會(huì)很難.