【答案】
分析:(1)把A,B的坐標(biāo)代入解析式,利用待定系數(shù)法即可求解;
(2)求得一次函數(shù)與x軸的交點(diǎn),根據(jù)S
△AOB=S
△AOC+S
△BOC即可求解;
(3)根據(jù)三角函數(shù)即可確定∠ACO=30°,判斷△OAC是底角為30°的等腰三角形,作QH⊥x軸,H為垂足,Rt△QOH中利用三角函數(shù)即可求得Q的坐標(biāo).
取OP
1=2OH=
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,則∠QP
1O=30°.過點(diǎn)Q作∠P
2QO=30°,交x軸于點(diǎn)P
2,則△OP
2Q∽△COA.根據(jù)雙曲線的對(duì)稱性,故可將△AOC繞原點(diǎn)O旋轉(zhuǎn)180°,得到△Q′O P
3,由此可得點(diǎn) Q′必在雙曲線左支上,點(diǎn)P
3在x軸正半軸上.即可求解.
解答:解:(1)∵一次函數(shù)與反比例函數(shù)相交于A、B兩點(diǎn),
∴
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與
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,∴
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,
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.(3分)
∴所求一次函數(shù)的解析式是
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,
所求反比例函數(shù)的解析式是
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.(4分)
(2)解法(一):由一次函數(shù)
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,令y=0,得x=-2.
∴點(diǎn)C的坐標(biāo)是(-2,0).(5分)
∴S
△AOB=S
△AOC+S
△BOC=
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(6分)=
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.(8分)
解法(二):分別過點(diǎn)A、B作AE⊥x軸于E,BF⊥y軸于F,且分別延長相交于G,
∴S
△AOB=S
△ABG-S
△BOF-S
△AOE-S
矩形OFGE=
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=
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(6分)=
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.(8分)
(3)設(shè)直線AC交y軸于點(diǎn)D,
∵
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,
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∴
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.
在Rt△COD中,
∵tan∠DCO=
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,
∴∠DCO=30°,即∠ACO=30°.
在Rt△AOE中,
∵tan∠AOE=
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,
∴∠AOE=60°.∴∠OAC=∠AOE-∠ACO=30°.
∴△OAC是底角為30°的等腰三角形.(9分)
作∠QOX=30°與反比例函數(shù)
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(x>0)的圖象交于點(diǎn)Q,設(shè)Q(
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),
作QH⊥x軸,H為垂足,
在Rt△QOH中,tan30°=
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,∴m
2=3,∴
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(取正數(shù))
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∴
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(10分)
取OP
1=2OH=
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,則∠QP
1O=30°.
∴△P
1QO∽△AOC.∴
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.(11分)
過點(diǎn)Q作∠P
2QO=30°,交x軸于點(diǎn)P
2,∴△OP
2Q∽△COA.
由∠QP
2H=60°,得
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,
∴P
2Q=
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.∴
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.(12分)
根據(jù)雙曲線的對(duì)稱性,故可將△AOC繞原點(diǎn)O旋轉(zhuǎn)180°,得到△Q′O P
3,由此可得點(diǎn) Q′必在雙曲線左支上,點(diǎn)P
3在x軸正半軸上.
∴
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,P
3(2,0).(14分)
綜上所述,所有符合條件的點(diǎn)的坐標(biāo)分別是
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,
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,P
3(2,0).
點(diǎn)評(píng):此題主要考查反比例函數(shù)的性質(zhì),注意通過解方程組求出交點(diǎn)坐標(biāo).同時(shí)要注意運(yùn)用數(shù)形結(jié)合的思想.