题目内容
(2002•烟台)如图,点A、B在反比例函数![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232530354024976/SYS201310212325303540249008_ST/0.png)
(1)求该反比例函数的解析式;
(2)若点(-a,y1),(-2a,y2)在该反比例函数的图象上,试比较y1与y2的大小;
(3)求△AOB的面积.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232530354024976/SYS201310212325303540249008_ST/images1.png)
【答案】分析:(1)由S△AOC=
xy=2,设反比例函数的解析式y=
,则k=xy=4;
(2)由于反比例函数的性质是:在x<0时,y随x的增大而减小,-a>-2a,则y1<y2;
(3)连接AB,过点B作BE⊥x轴,交x轴于E点,通过分割面积法S△AOB=S△AOC+S梯形ACEB-S△BOE求得.
解答:解:(1)∵S△AOC=2,
∴k=2S△AOC=4;
∴y=
;
(2)∵k>0,
∴函数y在各自象限内随x的增大而减小;
∵a>0,
∴-2a<-a;
∴y1<y2;
(3)连接AB,过点B作BE⊥x轴,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232530354024976/SYS201310212325303540249008_DA/images3.png)
S△AOC=S△BOE=2,
∴A(a,
),B(2a,
);
S梯形=
,
∴S△AOB=S△AOC+S梯形ACEB-S△BOE=3.
点评:此题重点检查函数性质的应用和图形的分割转化思想.同学们要熟练掌握这类题型.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232530354024976/SYS201310212325303540249008_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232530354024976/SYS201310212325303540249008_DA/1.png)
(2)由于反比例函数的性质是:在x<0时,y随x的增大而减小,-a>-2a,则y1<y2;
(3)连接AB,过点B作BE⊥x轴,交x轴于E点,通过分割面积法S△AOB=S△AOC+S梯形ACEB-S△BOE求得.
解答:解:(1)∵S△AOC=2,
∴k=2S△AOC=4;
∴y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232530354024976/SYS201310212325303540249008_DA/2.png)
(2)∵k>0,
∴函数y在各自象限内随x的增大而减小;
∵a>0,
∴-2a<-a;
∴y1<y2;
(3)连接AB,过点B作BE⊥x轴,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232530354024976/SYS201310212325303540249008_DA/images3.png)
S△AOC=S△BOE=2,
∴A(a,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232530354024976/SYS201310212325303540249008_DA/3.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232530354024976/SYS201310212325303540249008_DA/4.png)
S梯形=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021232530354024976/SYS201310212325303540249008_DA/5.png)
∴S△AOB=S△AOC+S梯形ACEB-S△BOE=3.
点评:此题重点检查函数性质的应用和图形的分割转化思想.同学们要熟练掌握这类题型.
![](http://thumb2018.1010pic.com/images/loading.gif)
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