![](http://thumb.1010pic.com/pic5/upload/201405/53623242656c0.png)
解:(1)如图,∵点A、B在x轴上(A点在B点左侧),A(-1,0),OB=3OA,
∴B(3,0).
又∵tan∠CAO=2,点C在y轴正半轴上,
∴
![](http://thumb.1010pic.com/pic5/latex/262523.png)
=2,则CO=2OA=2,
∴C(0,2)
综上所述,点B、C的坐标分别是:(3,0),(0,2);
(2)∵该抛物线与x轴的两个交点坐标是:A(-1,0),B(3,0),
∴设过点A、B、C的抛物线解析式为y=a(x+1)(x-3)(a≠0).
把点C的坐标代入,得
2=a(0+1)(0-3),
解得,a=-
![](http://thumb.1010pic.com/pic5/latex/168.png)
,
则该抛物线的解析式为:y=-
![](http://thumb.1010pic.com/pic5/latex/168.png)
(x+1)(x-3)(或y=-
![](http://thumb.1010pic.com/pic5/latex/168.png)
x
2+
![](http://thumb.1010pic.com/pic5/latex/304.png)
x+2);
(3)由(2)中抛物线解析式得到:y=-
![](http://thumb.1010pic.com/pic5/latex/168.png)
(x-1)
2+
![](http://thumb.1010pic.com/pic5/latex/137.png)
,则顶点P的坐标是(1,
![](http://thumb.1010pic.com/pic5/latex/137.png)
).
∵△ABQ与△ABP的面积相等,且点Q是抛物线上的一点
∴点Q与点P到x轴的距离相等,
∴点Q是直线y=±
![](http://thumb.1010pic.com/pic5/latex/137.png)
与抛物线的交点.
①当y=
![](http://thumb.1010pic.com/pic5/latex/137.png)
时,x=1,此时,点Q与点P重合,即Q(1,
![](http://thumb.1010pic.com/pic5/latex/137.png)
);
②当y=-
![](http://thumb.1010pic.com/pic5/latex/137.png)
时,-
![](http://thumb.1010pic.com/pic5/latex/168.png)
(x-1)
2+
![](http://thumb.1010pic.com/pic5/latex/137.png)
=-
![](http://thumb.1010pic.com/pic5/latex/137.png)
,
解得,x
1=1+2
![](http://thumb.1010pic.com/pic5/latex/53.png)
,x
2=1-2
![](http://thumb.1010pic.com/pic5/latex/53.png)
,此时,点Q的坐标是(1+2
![](http://thumb.1010pic.com/pic5/latex/53.png)
,-
![](http://thumb.1010pic.com/pic5/latex/137.png)
)或(1-2
![](http://thumb.1010pic.com/pic5/latex/53.png)
,-
![](http://thumb.1010pic.com/pic5/latex/137.png)
)
综上所述,符合条件的点Q的坐标是:(1,
![](http://thumb.1010pic.com/pic5/latex/137.png)
)、(1+2
![](http://thumb.1010pic.com/pic5/latex/53.png)
,-
![](http://thumb.1010pic.com/pic5/latex/137.png)
)或(1-2
![](http://thumb.1010pic.com/pic5/latex/53.png)
,-
![](http://thumb.1010pic.com/pic5/latex/137.png)
).
分析:(1)根据已知条件“A(-1,0),OB=3OA,且tan∠CAO=2”易求点B、C的坐标;
(2)设抛物线解析式为y=a(x+1)(x-3)(a≠0).然后把点C的坐标分别代入,求得a的值;
(3)根据“同底等高的两个三角形的面积相等”可知,点Q是直线y=与抛物线的交点.
点评:本题考查了二次函数综合题.其中涉及到了坐标与图形的性质,锐角三角函数的定义,待定系数法求二次函数的解析式,以及一次函数与抛物线的交点问题.解答(2)题时,因为已知抛物线与x轴的两个交点坐标,所以设交点式关系式,可以减少繁琐的计算过程.