试题分析:(1)利用相似及相似比,可得到C的坐标.把A,B代入一次函数解析式即可求得解析式的坐标.
(2)顶点落在x轴正半轴上说明此函数解析式与x轴有一个交点,那么△=0,再把B,C两点即可.
(3)到直线AB的距离为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451345433.png)
的直线有两条,可求出这两条直线解析式,和二次函数解析式组成方程组,求得点P坐标.
(1)过C点向x轴作垂线,垂足为D,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082302445151727537.png)
由位似图形性质可知:△ABO∽△ACD,
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451532909.png)
.
由已知
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451548562.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451548550.png)
可知:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451563619.png)
.
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451579595.png)
.
∴C点坐标为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451361484.png)
.
设直线BC的解析式为: y=kx+4,将(5,9)代入得5k+4=9,解得k=1.
所以y=x+4.
(2)因为抛物线顶点在x轴正半轴,所以设顶点坐标为(h,0),则设抛物线解析式为y=a(x-h)
2.
将(0,4),(5,9)代入函数解析式得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408230244516411032.png)
,解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451641691.png)
或者
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451657879.png)
.
∴解得抛物线解析式为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451673642.png)
或
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451688900.png)
.
又∵
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451688900.png)
的顶点在x轴负半轴上,不合题意,故舍去.
∴满足条件的抛物线解析式为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451423617.png)
(准确画出函数
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451423617.png)
图象)
(3)将直线BC绕B点旋转与抛物线相交与另一点P,设P到 直线AB的距离为h,
故P点应在与直线AB平行,且相距
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451345433.png)
的上下两条平行直线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451766319.png)
和
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451782343.png)
上.
由平行线的性质可得:两条平行直线与y轴的交点到直线BC的距离也为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451345433.png)
.
设
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451766319.png)
与y轴交于E点,过E作EF⊥BC于F点,
在Rt△BEF中
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451829704.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451860888.png)
,
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451875496.png)
.
∴可以求得直线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451766319.png)
与y轴交点坐标为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451907470.png)
同理可求得直线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451782343.png)
与y轴交点坐标为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451938508.png)
∴两直线解析式
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451953628.png)
;
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451969622.png)
.
根据题意列出方程组:(1)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451985932.png)
;(2)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024452000905.png)
∴解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024452016771.png)
;
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024452031762.png)
;
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024452063775.png)
;
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024452078736.png)
∴满足条件的点P有四个,它们分别是
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451439573.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451454572.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451470593.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823024451485554.png)
.
点评:此类问题综合性强,难度较大,在中考中比较常见,一般作为压轴题,题目比较典型.