试题分析:(1)∵
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002337850302.png)
=x
2,∴在点P(2,4)处的切线的斜率k=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002337850302.png)
|
x=2="4." ……………2分
∴曲线在点P(2,4)处的切线方程为y-4=4(x-2),即4x-y-4="0." …………………… 4分
(2)设曲线y=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002337881531.png)
与过点P(2,4)的切线相切于点
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002337897861.png)
,
则切线的斜率k=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002337850302.png)
|
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002337928228.png)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002337928291.png)
. ……………… 6分
∴切线方程为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240023379431080.png)
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002337959716.png)
…………………… 8分
∵点P(2,4)在切线上,∴4=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002337975722.png)
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002338021882.png)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002338037748.png)
∴(x
0+1)(x
0-2)
2=0,解得x
0=-1或x
0=2,
故所求的切线方程为4x-y-4=0或x-y+2=0. ……………………12分
点评:易错题,求曲线的切线问题,往往包括两种类型,一是知切点,二是过曲线外的点,后者难度大些。