【答案】
分析:(1)由BD∥y軸,可知B點(diǎn)與D點(diǎn)的橫坐標(biāo)相等,將x=-8代入直線y=
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x,即可求出點(diǎn)B的坐標(biāo);再根據(jù)A點(diǎn)與B點(diǎn)關(guān)于原點(diǎn)對(duì)稱(chēng),求出A點(diǎn)坐標(biāo);
(2)先由B是CD中點(diǎn),D點(diǎn)縱坐標(biāo)為0,可知B點(diǎn)縱坐標(biāo)是C點(diǎn)縱坐標(biāo)的
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,即為-
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,又B點(diǎn)在直線y=
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x上,把y=-
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代入直線y=
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x,得B點(diǎn)橫坐標(biāo)為-2n,從而可用含n的代數(shù)式表示k及E點(diǎn)的坐標(biāo),然后根據(jù)四邊形OBCE的面積=矩形ODCN面積-直角三角形ODB的面積-直角三角形ONE的面積,列出關(guān)于n的方程,解方程求出n的值,即可得出C、M兩點(diǎn)的坐標(biāo),最后運(yùn)用待定系數(shù)法求出直線CM的解析式;
(3)由于點(diǎn)M(m,n)在雙曲線
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上,得出k=mn,再聯(lián)立雙曲線y=
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與直線y=
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x,求出A、B兩點(diǎn)的坐標(biāo),由MA=pMP,MB=qMQ求出p、q,從而得出p-q的值.
解答:解:(1)將x=-8代入直線y=
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x,
得y=-2.
∴點(diǎn)B坐標(biāo)(-8,-2),--(1分)
將點(diǎn)B坐標(biāo)(-8,-2)代入
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得:
k=xy=16.--(2分)
∵A點(diǎn)是B點(diǎn)關(guān)于原點(diǎn)的對(duì)稱(chēng)點(diǎn),
∴A點(diǎn)坐標(biāo)為(8,2).--(3分)
(2)∵B是CD中點(diǎn),C點(diǎn)縱坐標(biāo)為-n,
∴B點(diǎn)縱坐標(biāo)為-
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,
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把y=-
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代入直線y=
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x,得B點(diǎn)橫坐標(biāo)為-2n,
∴D點(diǎn)坐標(biāo)(-2n,0),B點(diǎn)坐標(biāo)(-2n,-
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),C點(diǎn)坐標(biāo)(-2n,-n).--(4分)
∴k=(-2n)×(-
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)=n
2.
將E點(diǎn)縱坐標(biāo)-n代入方程y=n
2/x,得其橫坐標(biāo)-n.
∵四邊形OBCE的面積=矩形ODCN面積-Rt△ODB的面積-Rt△ONE的面積,
∴4=2n
2-
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n
2-
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n
2,
解得n=2.--(5分)
所以C點(diǎn)坐標(biāo)(-4,-2),M點(diǎn)坐標(biāo)(2,2)--(6分)
設(shè)直線CM的解析式為y=kx+b,則
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,
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解得
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.
∴直線CM解析式為y=
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x+
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.--(7分)
(3)將點(diǎn)M的坐標(biāo)(m,n)代入雙曲線方程得:k=mn.
雙曲線y=
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與直線y=
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x聯(lián)立,
解得A點(diǎn)坐標(biāo)(2
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,
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),B點(diǎn)坐標(biāo)(-2
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,-
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),
∴MA=
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,
MP=
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,
∵M(jìn)A=pMP,MB=qMQ,
∴p=
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=
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,--(9分)
q=
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=
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,--(11分)
∴p-q=
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-
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=-2.--(12分)
點(diǎn)評(píng):此題綜合考查了反比例函數(shù),正比例函數(shù)等多個(gè)知識(shí)點(diǎn).此題難度稍大,綜合性比較強(qiáng),注意對(duì)各個(gè)知識(shí)點(diǎn)的靈活應(yīng)用.