解:(1)由抛物线的轴对称性可知, C、F两点关于抛物线对称轴y轴对称, ∴C为(-2,2), 而抛物线的顶点为(0,1), 设解析式为y=ax2+1, 把点(-2,2)代入得 , ∴ ; (2)PS=PB, 理由如下: 设P为 , ∴![](http://img.shitiku.com.cn/uploads/allimg/20191020/20191020174228-39012.gif) PB=![](http://img.shitiku.com.cn/uploads/allimg/20191020/20191020174228-50486.gif) ∴PS=PB; (3)设SP=b,QR=c, 由(2)可知SP=PB=b,QR=BQ=c, ∴SR=![](http://img.shitiku.com.cn/uploads/allimg/20191020/20191020174228-52308.gif)
即![](http://img.shitiku.com.cn/uploads/allimg/20191020/20191020174229-46063.gif) 设SR上有一点M, ∴①当△SPM∽RQM时
![](http://img.shitiku.com.cn/uploads/allimg/20191020/20191020174229-10815.gif) ∴![](http://img.shitiku.com.cn/uploads/allimg/20191020/20191020174229-67720.gif) ∴M与原点O重合, ②当△SPM∽△RMQ时,![](http://img.shitiku.com.cn/uploads/allimg/20191020/20191020174229-15579.gif) 即![](http://img.shitiku.com.cn/uploads/allimg/20191020/20191020174230-83030.gif) ∴![](http://img.shitiku.com.cn/uploads/allimg/20191020/20191020174230-72819.gif) 解得![](http://img.shitiku.com.cn/uploads/allimg/20191020/20191020174230-57610.gif) ∴M为SR的中点, 综上所述:M为SR的中点或与原点重合。 |