已知点A（m、n）是反比例函数（x＞0）的图象上一点，过A作AB⊥x轴于点B，P是y轴上一点，        ...

问题详情：

已知点A（m、n）是反比例函数（x＞0）的图象上一点，过A作AB⊥x轴于点B，P是y轴上一点，                                                       

（1）求△PAB的面积；                                                                      

（2）当△PAB为等腰直角三角形时，求点A的坐标；                                     

（3）若∠APB=90°，求m的取值范围．                                              

【回答】

解：（1）连接OA，                                                                          

∵AB⊥x轴，                                                                                     

∴AB∥y轴，                                                                                     

∴S△PAB=S△POB，                                                                              

∵点A（m、n）是反比例函数（x＞0）的图象上一点，                            

∴S△PAB=S△POB=2；                                                                          

（2）若∠ABP=90°，则AB=OB，                                                        

则m=n，                                                                                           

∴m=，                                                                                           

∵x＞0，                                                                                            

∴m=2，                                                                                            

∴点A（2，2）；                                                                              

若∠PAB=90°，则PA=AB，同理可得点A（2，2）；                                       

若∠APB=90°，则AP=BP，                                                                 

过点P作PC⊥AB于点C，则AC=BC=PC，                                         

则点A（m，2m），                                                                          

∴2m=，                                                                                          

∵x＞0，                                                                                            

∴m=，                                                                                         

∴点A（，2）；                                                                       

综上，点A的坐标为：（2，2）或（，2）；                                         

（3）∵∠APB=90°，                                                                         

∴点P是以AB为直径的圆与y轴的交点，                                          

由（2）可知当x=时，以AB为直径的圆与y轴相切，当x＞时，以AB为直径的圆与y轴相离，                                                           

∴m的取值范围为：0＜m≤．                                                         

知识点：反比例函数

题型：解答题