平面直角坐標(biāo)系中,過原點O的直線l與曲線y=ex-1交于不同的A,B兩點,分別過點A,B作y軸的平行線,與曲線y=lnx交于點C,D,則直線CD的斜率是 .
【答案】
分析:設(shè)直線l的方程為y=kx(k>0),A(x
1,y
1),B(x
2,y
2)(x
1>0,x
2>0),則直線CD的斜率k
CD=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103323477060911/SYS201311031033234770609016_DA/0.png)
,根據(jù)A、B為直線l與曲線y=e
x-1交點可得kx
1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103323477060911/SYS201311031033234770609016_DA/1.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103323477060911/SYS201311031033234770609016_DA/2.png)
,兩邊取對數(shù)后代人斜率公式即可求得答案.
解答:解:設(shè)直線l的方程為y=kx(k>0),A(x
1,y
1),B(x
2,y
2)(x
1>0,x
2>0),
則C(x
1,lnx
1),D(x
2,lnx
2),
所以kx
1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103323477060911/SYS201311031033234770609016_DA/3.png)
⇒x
1-1=lnkx
1,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103323477060911/SYS201311031033234770609016_DA/4.png)
⇒x
2-1=lnkx
2,
所以直線CD的斜率k
CD=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103323477060911/SYS201311031033234770609016_DA/5.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103323477060911/SYS201311031033234770609016_DA/6.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103323477060911/SYS201311031033234770609016_DA/7.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103323477060911/SYS201311031033234770609016_DA/8.png)
=1,
故答案為:1.
點評:本題考查直線與圓錐曲線的位置關(guān)系及直線斜率的求解,考查數(shù)形結(jié)合思想及學(xué)生計算能力、解決問題的能力.