已知
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021216977333.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217008416.png)
,曲線
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217023313.png)
上任意一點(diǎn)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217023289.png)
分別與點(diǎn)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217070553.png)
、
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217086536.png)
連線的斜率的乘積為
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217101486.png)
.
(Ⅰ)求曲線
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217023313.png)
的方程;
(Ⅱ)設(shè)直線
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217133949.png)
與
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217133276.png)
軸、
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217148308.png)
軸分別交于
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217164399.png)
、
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217179357.png)
兩點(diǎn),若曲線
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217023313.png)
與直線
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217211271.png)
沒有公共點(diǎn),求證:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217226765.png)
.
(Ⅰ)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217320762.png)
,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217335432.png)
.
(Ⅱ)由
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240212173511200.png)
得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240212173671263.png)
,利用曲線
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217023313.png)
與直線
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217211271.png)
沒有公共點(diǎn),
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217398420.png)
,得到
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217413700.png)
,利用
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217429799.png)
,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217445620.png)
,及均值定理確定
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240212174602744.png)
,
從而證得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217476762.png)
.
試題分析:(Ⅰ)設(shè)曲線
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217023313.png)
上任意一點(diǎn)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217023289.png)
的坐標(biāo)為
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217523500.png)
.利用依題意點(diǎn)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217023289.png)
分別與點(diǎn)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217070553.png)
、
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217086536.png)
連線的斜率的乘積為
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217101486.png)
,轉(zhuǎn)化成代數(shù)式,整理可得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217320762.png)
.
(Ⅱ)由
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240212173511200.png)
得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240212176161307.png)
,利用曲線
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217023313.png)
與直線
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217211271.png)
沒有公共點(diǎn),
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217398420.png)
,得到
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217413700.png)
,利用
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217429799.png)
,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217445620.png)
,及均值定理確定
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240212177252699.png)
,
從而證得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217476762.png)
.
試題解析:(Ⅰ)設(shè)曲線
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217023313.png)
上任意一點(diǎn)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217023289.png)
的坐標(biāo)為
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217523500.png)
.
依題意
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240212178191147.png)
,且
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217335432.png)
, 3分
整理得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217320762.png)
.所以,曲線
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217023313.png)
的方程為:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217320762.png)
,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217335432.png)
. 5分
(Ⅱ)由
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240212173511200.png)
得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240212179281252.png)
,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240212179442050.png)
, 7分
由已知條件可知
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217429799.png)
,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217445620.png)
,所以
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240212177252699.png)
,
從而
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217991910.png)
, 即
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021217476762.png)
. 13分
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,
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分別是雙曲線
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021153655339.png)
:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021153670749.png)
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:
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的一個(gè)交點(diǎn)為
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021153764289.png)
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