如图，反比例函数y=的图象与一次函数y=x的图象交于点A、B，点B的横坐标是4．点P是第一象限内反比例函数图象...

问题详情：

如图，反比例函数y=的图象与一次函数y=x的图象交于点A、B，点B的横坐标是4．点P是第一象限内反比例函数图象上的动点，且在直线AB的上方．                                                    

（1）若点P的坐标是（1，4），直接写出k的值和△PAB的面积；                

（2）设直线PA、PB与x轴分别交于点M、N，求*：△PMN是等腰三角形；                                   

（3）设点Q是反比例函数图象上位于P、B之间的动点（与点P、B不重合），连接AQ、BQ，比较∠PAQ与∠PBQ的大小，并说明理由．                                                                              

【回答】

【解答】解：（1）k=4，S△PAB=15．                                                  

提示：过点A作AR⊥y轴于R，过点P作PS⊥y轴于S，连接PO，                

设AP与y轴交于点C，如图1，                                                          

把x=4代入y=x，得到点B的坐标为（4，1），                                            

把点B（4，1）代入y=，得k=4．                                                     

解方程组，得到点A的坐标为（﹣4，﹣1），                                     

则点A与点B关于原点对称，                                                             

∴OA=OB，                                                                                       

∴S△AOP=S△BOP，                                                                              

∴S△PAB=2S△AOP．                                                                            

设直线AP的解析式为y=mx+n，                                                         

把点A（﹣4，﹣1）、P（1，4）代入y=mx+n，                                             

求得直线AP的解析式为y=x+3，                                                        

则点C的坐标（0，3），OC=3，                                                        

∴S△AOP=S△AOC+S△POC                                                                       

=OCAR+OCPS                                                                               

=×3×4+×3×1=，                                                                        

∴S△PAB=2S△AOP=15；                                                                        

（2）过点P作PH⊥x轴于H，如图2．                                               

B（4，1），则反比例函数解析式为y=，                                          

设P（m，），直线PA的方程为y=ax+b，直线PB的方程为y=px+q，                                  

联立，解得直线PA的方程为y=x+﹣1，                              

联立，解得直线PB的方程为y=﹣x++1，                                    

∴M（m﹣4，0），N（m+4，0），                                                    

∴H（m，0），                                                                                 

∴MH=m﹣（m﹣4）=4，NH=m+4﹣m=4，                                         

∴MH=NH，                                                                                       

∴PH垂直平分MN，                                                                         

∴PM=PN，                                                                                       

∴△PMN是等腰三角形；                                                                   

（3）∠PAQ=∠PBQ．                                                                       

理由如下：                                                                                        

过点Q作QT⊥x轴于T，设AQ交x轴于D，QB的延长线交x轴于E，如图3．                                 

可设点Q为（c，），直线AQ的解析式为y=px+q，则有                              

，                                                                               

解得：，                                                                            

∴直线AQ的解析式为y=x+﹣1．                                                   

当y=0时， x+﹣1=0，                                                                   

解得：x=c﹣4，                                                                                 

∴D（c﹣4，0）．                                                                             

同理可得E（c+4，0），                                                                     

∴DT=c﹣（c﹣4）=4，ET=c+4﹣c=4，                                              

∴DT=ET，                                                                                        

∴QT垂直平分DE，                                                                          

∴QD=QE，                                                                                       

∴∠QDE=∠QED．                                                                            

∵∠MDA=∠QDE，                                                                           

∴∠MDA=∠QED．                                                                           

∵PM=PN，∴∠PMN=∠PNM．                                                          

∵∠PAQ=∠PMN﹣∠MDA，∠PBQ=∠NBE=∠PNM﹣∠QED，                      

∴∠PAQ=∠PBQ．                                                                            

知识点：反比例函数

题型：综合题