题目内容
(2006•成都)如图,已知反比例函数y=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_ST/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_ST/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_ST/2.png)
(1)求k和m的值;
(2)若一次函数y=ax+1的图象经过点A,并且与x轴相交于点C,求∠ACO的度数和|AO|:|AC|的值.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_ST/images3.png)
【答案】分析:(1)根据△AOB的面积为
,得到反比例函数的解析式,进而可以求出m的值.
(2)把A(-
,2)代入y=ax+1中,就可以求出a的值,得到函数的解析式,因而求出C点的坐标,在Rt△ABC中就可以求出tan∠ACO的值,得到AC的值,在Rt△ABO中,根据勾股定理就可以求出OA的值.
解答:解:(1)∵k<0,
∴点A(-
,m)在第二象限内.
∴m>0,|OB|=|-
|=
,|AB|=m.
∵S△AOB=
•|OB|•|AB|=
•
•m=
,
∴m=2.
∴点A的坐标为A(-
,2).(2分)
把A(-
,2)的坐标代入y=
中,
得2=
,
∴k=-2
.(2分)
(2)把A(-
,2)代入y=ax+1中,得2=-
a+1,
∴a=
.
∴y=-
.(1分)
令y=0,得-
x+1=0,
∴x=
.
∴点C的坐标为C(
,0).
∵AB⊥x轴于点B,
∴△ABC为直角三角形.
在Rt△ABC中,|AB|=2,|BC|=2
,
∴tan∠ACO=
,
∴∠ACO=30°.
∴|AC|=2|AB|=4.(2分)
在Rt△ABO中,由勾股定理,
得|AO|=
.
∴|AO|:|AC|=
:4.(1分)
点评:本题考查函数图象交点坐标的求法及反比例函数的比例系数k与其图象上的点与原点所连的线段、坐标轴、向坐标轴作垂线所围成的直角三角形面积S的关系,即S=
|k|.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/0.png)
(2)把A(-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/1.png)
解答:解:(1)∵k<0,
∴点A(-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/2.png)
∴m>0,|OB|=|-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/3.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/4.png)
∵S△AOB=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/5.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/6.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/7.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/8.png)
∴m=2.
∴点A的坐标为A(-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/9.png)
把A(-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/10.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/11.png)
得2=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/12.png)
∴k=-2
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/13.png)
(2)把A(-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/14.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/15.png)
∴a=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/16.png)
∴y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/17.png)
令y=0,得-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/18.png)
∴x=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/19.png)
∴点C的坐标为C(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/20.png)
∵AB⊥x轴于点B,
∴△ABC为直角三角形.
在Rt△ABC中,|AB|=2,|BC|=2
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/21.png)
∴tan∠ACO=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/22.png)
∴∠ACO=30°.
∴|AC|=2|AB|=4.(2分)
在Rt△ABO中,由勾股定理,
得|AO|=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/23.png)
∴|AO|:|AC|=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/24.png)
点评:本题考查函数图象交点坐标的求法及反比例函数的比例系数k与其图象上的点与原点所连的线段、坐标轴、向坐标轴作垂线所围成的直角三角形面积S的关系,即S=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904444183672/SYS201310212319044441836005_DA/25.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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