如图，点A是双曲线y=在第一象限上的一动点，连接AO并延长交另一分支于点B，以AB为斜边作等腰Rt△ABC，点...

问题详情：

如图，点A是双曲线y=在第一象限上的一动点，连接AO并延长交另一分支于点B，以AB为斜边作等腰Rt△ABC，点C在第二象限，随着点A的运动，点C的位置也不断的变化，但始终在一函数图象上运动，则这个函数的解析式为　　．                                                                                                          

【回答】

y=﹣【解答】解：连结OC，作CD⊥x轴于D，AE⊥x轴于E，如图，                                  

设A点坐标为（a，），                                                                    

∵A点、B点是正比例函数图象与双曲线y=的交点，                                    

∴点A与点B关于原点对称，                                                             

∴OA=OB                                                                                           

∵△ABC为等腰直角三角形，                                                             

∴OC=OA，OC⊥OA，                                                                       

∴∠DOC+∠AOE=90°，                                                                      

∵∠DOC+∠DCO=90°，                                                                      

∴∠DCO=∠AOE，                                                                            

∵在△COD和△OAE中                                                                       

∴△COD≌△OAE（AAS），                                                              

∴OD=AE=，CD=OE=a，                                                                  

∴C点坐标为（﹣，a），                                                                 

∵﹣a=﹣4，                                                                                    

∴点C在反比例函数y=﹣图象上．                                                   

故*为y=﹣．                                                                              

知识点：反比例函数

题型：填空题