如圖，反比例函式y=的圖象與一次函式y=x的圖象交於點A、B，點B的橫座標是4．點P是第一象限內反比例函式圖象...

問題詳情：

如圖，反比例函式y=的圖象與一次函式y=x的圖象交於點A、B，點B的橫座標是4．點P是第一象限內反比例函式圖象上的動點，且在直線AB的上方．                                                    

（1）若點P的座標是（1，4），直接寫出k的值和△PAB的面積；                

（2）設直線PA、PB與x軸分別交於點M、N，求*：△PMN是等腰三角形；                                   

（3）設點Q是反比例函式圖象上位於P、B之間的動點（與點P、B不重合），連線AQ、BQ，比較∠PAQ與∠PBQ的大小，並說明理由．                                                                              

【回答】

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

提示：過點A作AR⊥y軸於R，過點P作PS⊥y軸於S，連線PO，                

設AP與y軸交於點C，如圖1，                                                          

把x=4代入y=x，得到點B的座標為（4，1），                                            

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

解方程組，得到點A的座標為（﹣4，﹣1），                                     

則點A與點B關於原點對稱，                                                             

∴OA=OB，                                                                                       

∴S△AOP=S△BOP，                                                                              

∴S△PAB=2S△AOP．                                                                            

設直線AP的解析式為y=mx+n，                                                         

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

求得直線AP的解析式為y=x+3，                                                        

則點C的座標（0，3），OC=3，                                                        

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

=OCAR+OCPS                                                                               

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

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

（2）過點P作PH⊥x軸於H，如圖2．                                               

B（4，1），則反比例函式解析式為y=，                                          

設P（m，），直線PA的方程為y=ax+b，直線PB的方程為y=px+q，                                  

聯立，解得直線PA的方程為y=x+﹣1，                              

聯立，解得直線PB的方程為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：                                                                                        

過點Q作QT⊥x軸於T，設AQ交x軸於D，QB的延長線交x軸於E，如圖3．                                 

可設點Q為（c，），直線AQ的解析式為y=px+q，則有                              

，                                                                               

解得：，                                                                            

∴直線AQ的解析式為y=x+﹣1．                                                   

當y=0時， 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．                                                                            

知識點：反比例函式

題型：綜合題