![](http://thumb.1010pic.com/pic5/upload/201310/5286b866a165d.png)
解:(1)过点H作DH⊥CO,
∵点C在x轴的正半轴上且坐标为(4,O),△ODC是以CO为斜边的等腰直角三角形,
∴DH=HO=HC=2,
∴由题可知:D(2,2),
∵点D在反比例函数y=
![](http://thumb.1010pic.com/pic5/latex/9448.png)
(k>0)上,
∴k=2×2=4;
(2)连接DD
1,CD
1,
∵线段D
1C
1,由线段DC平移而成,
∴四边形DD
1C
1C为平行四边形,
∴D
1于点C关于原点对称,
∵C(4,0),
∴D
1(-4,0);
(3)∵点A(a,m),B(3a,b)在反比例函数y=
![](http://thumb.1010pic.com/pic5/latex/2698.png)
(k>0)的图象上,
∴am=3ab,即b=
![](http://thumb.1010pic.com/pic5/latex/8663.png)
,am=4,
分别过点AB作AC⊥x轴,BD⊥x轴,垂足分别为C、D,
∴S
△AOC=S
△BOD=
![](http://thumb.1010pic.com/pic5/latex/13.png)
k=
![](http://thumb.1010pic.com/pic5/latex/13.png)
×4=2,
∴S
△OAB=S
梯形ACDB,即S
△OAB=
![](http://thumb.1010pic.com/pic5/latex/13.png)
(m+b)×(3a-a)=
![](http://thumb.1010pic.com/pic5/latex/13.png)
×
![](http://thumb.1010pic.com/pic5/latex/304.png)
m×2a=
![](http://thumb.1010pic.com/pic5/latex/260636.png)
=
![](http://thumb.1010pic.com/pic5/latex/260637.png)
=
![](http://thumb.1010pic.com/pic5/latex/5689.png)
.
分析:(1)由于△OCD是等腰直角三角形,不难得出D(2,2),将其代入反比例函数的解析式y=
![](http://thumb.1010pic.com/pic5/latex/9448.png)
(k>0)中即可求出k的值;
(2)连接DD
1,CD
1可知四边形DD
1C
1C为平行四边形,根据平行四边形的性质即可得出点D
1的坐标;
(3)先根据动点A(a,m),B(3a,b)在反比例函数y=
![](http://thumb.1010pic.com/pic5/latex/9448.png)
(k>0)的图象上,故am=3ab,即b=
![](http://thumb.1010pic.com/pic5/latex/8663.png)
,分别过点AB作AC⊥x轴,BD⊥x轴,垂足分别为C、D,由反比例函数系数k的几何意义可知,S
△AOC=S
△BOD=
![](http://thumb.1010pic.com/pic5/latex/13.png)
k=2,故S
△OAB=S
梯形ACDB,由此即可得出结论.
点评:本题考查的是反比例函数综合题,熟知反比例函数系数k的几何意义及反比例函数图象上点的坐标特点是解答此题的关键.