【答案】
分析:(1)因?yàn)橹本€l
1經(jīng)過點(diǎn)A(-2,0)和點(diǎn)B(0,
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),可設(shè)直線l
1的解析式為y=kx+b,將A、B的坐標(biāo)代入,利用方程組即可求得該解析式,又因直線l
2的函數(shù)表達(dá)式為y=-
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x+
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,l
1與l
2相交于點(diǎn)P,所以將兩函數(shù)解析式聯(lián)立得到方程組,解之即可得到交點(diǎn)P的坐標(biāo),過P作x軸的垂線段,垂足為H,由P的坐標(biāo)可知,AH=EH=3,PH=
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,所以∠PEA=∠PAE=30°,利用三角形的外角等于和它不相鄰的兩個(gè)內(nèi)角的和,可得到∠FPB是60°;
(2)當(dāng)C在射線PA的延長線上時(shí),設(shè)⊙C和直線l
2相切時(shí),D是切點(diǎn),連接CD,則CD⊥PD.過點(diǎn)P作CM的垂線PG,垂足為G,因?yàn)椤螩PG=∠CAB=30°,PC=PC,所以可證Rt△CDP≌Rt△PGC,所以PG=CD=R.當(dāng)點(diǎn)C在射線PA上,⊙C和直線l
2相切時(shí),同理可證Rt△CDP≌Rt△PGC,所以PG=CD=R.取R=
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-2時(shí),C在AP的反向延長線上時(shí),因?yàn)镻的橫坐標(biāo)為1,所以a=1+R=
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-1,C在PA上時(shí),因?yàn)镻的橫坐標(biāo)為1,所以a=-(R-1)=3-3
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;
(3)當(dāng)⊙C和直線l
2不相離時(shí),由(2)知,分兩種情況討論:①當(dāng)0≤a≤
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-1時(shí),四邊形是一個(gè)直角梯形,所以有
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=
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,利用二次函數(shù)的頂點(diǎn)公式即可求出S的最大值;
②當(dāng)=3-3
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≤a<0時(shí),顯然⊙C和直線l
2相切即a=3-3
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時(shí),S最大.此時(shí)
s
最大值=
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,綜合以上①和②,即可求出答案.
解答:
解:(1)設(shè)直線l
1的解析式為y=kx+b,
∵直線l
1經(jīng)過點(diǎn)A(-2,0)和點(diǎn)B(0,
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),
∴
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,
解得
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,
∴直線l
1的解析式為y=
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x+
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;
聯(lián)立l
1與l
2得,
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,
解得
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,
∴P(1,
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);
過P作x軸的垂線段,垂足為H,
∵P(1,
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),
∴AH=EH=3,PH=
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,
∴∠PEA=∠PAE=30°,
∴∠FPB=∠PEA+∠PAE=60°;
(2)設(shè)⊙C和直線l
2相切時(shí)的一種情況如圖1所示,D是切點(diǎn),連接CD,則CD⊥PD.過點(diǎn)P作CM的垂線PG,垂足為G,則Rt△CDP≌Rt△PGC(∠PCD=∠CPG=30°,CP=PC),
∴PG=CD=R.
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當(dāng)點(diǎn)C在射線PA上,⊙C和直線l
2相切時(shí),同理可證.
取R=
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-2時(shí),a=1+R=
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-1,或a=-(R-1)=3-3
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;
(3)當(dāng)⊙C和直線l
2不相離時(shí),由(2)知,分兩種情況討論:
①如圖2,當(dāng)0≤a≤
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-1時(shí),
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=
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,
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時(shí),(滿足a≤
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-1),S有最大值.此時(shí)s
最大值=
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(
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);
②當(dāng)=3-3
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≤a<0時(shí),顯然⊙C和直線l
2相切即a=3-3
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時(shí),S最大.此時(shí)
s
最大值=
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.
綜合以上①和②,當(dāng)a=3或a=
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時(shí),存在S的最大值,其最大面積為
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.
點(diǎn)評:考查一次函數(shù)的解析式、圖象、性質(zhì)和圓的相關(guān)知識,及綜合應(yīng)用相關(guān)知識分析問題、解決問題的能力.
此題也較為新穎,符合新課標(biāo)的理念,揭示了求最值的一般方法,本題的難度設(shè)置也較為合適,使同學(xué)們都能有發(fā)揮自己能力的空間.