试题分析:解:根据题意,A(a,0)、B(0,b)、C(-2,-2)三点共线,可得k
AB=k
BC,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824013545266732.png)
,化简可得2a+2b+ab=0,即ab=-2a-2b,若ab>0,要么a>0且b>0,要么a<0且b<0,直线经过第三象限的C(-2,-2),由直线的性质可知,a<0,b<0,因为a<0,b<0,所以-2a-2b>0且-2a-2b≥2
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824013545297498.png)
=4
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824013545313429.png)
又因为ab=-2a-2b,所以ab≥4
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824013545313429.png)
,即ab-4
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824013545313429.png)
≥0,令t=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824013545313429.png)
>0,可得t
2-4t≥0,解可得t≥4或t≤0,又由t>0,则t≥4,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824013545313429.png)
≥4,ab≥16;则ab的最小值为16;故答案为16.
点评:本题考查基本不等式的应用,涉及三点共线的问题,有一定的难度;解题的难点在于利用基本不等式对(-2a-2b)变形,可得ab≥4
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824013545313429.png)
,进而由一元二次不等式的性质来求解