如圖1，已知直線y=2x分別與雙曲線y=、y=（x＞0）交於P、Q兩點，且OP=2OQ．           ...

問題詳情：

如圖1，已知直線y=2x分別與雙曲線y=、y=（x＞0）交於P、Q兩點，且OP=2OQ．                  

（1）求k的值．                                                                                

（2）如圖2，若點A是雙曲線y=上的動點，AB∥x軸，AC∥y軸，分別交雙曲線y=（x＞0）於點B、C，連線BC．請你探索在點A運動過程中，△ABC的面積是否變化？若不變，請求出△ABC的面積；若改變，請說明理由；                

（3）如圖3，若點D是直線y=2x上的一點，請你進一步探索在點A運動過程中，以點A、B、C、D為頂點的四邊形能否為平行四邊形？若能，求出此時點A的座標；若不能，請說明理由．                

【回答】

【解答】解：（1）過點Q作QE⊥x軸，垂足為E，過點P作PF⊥x軸，垂足為F，如圖1，                   

聯立，                                                                                    

解得：或．                                                                        

∵x＞0，                                                                                            

∴點P的座標為（2，4）．                                                                     

∴OF=2，PF=4．                                                                               

∵QE⊥x軸，PF⊥x軸，                                                                          

∴QE∥PF．                                                                                       

∴△OEQ∽△OFP．                                                                           

∴==．                                                                                 

∵OP=2OQ，                                                                                     

∴OF=2OE=2，PF=2EQ=4．                                                                    

∴OE=1，EQ=2．                                                                               

∴點Q的座標為（1，2）．                                                                     

∵點Q（1，2）在雙曲線y=上，                                                           

∴k=1×2=2．                                                                                      

∴k的值為2．                                                                                   

（2）如圖2，                                                                                    

設點A的座標為（a，b），                                                                     

∵點A（a，b）在雙曲線y=上，                                                           

∴b=．                                                                                            

∵．AB∥x軸，AC∥y軸，                                                                     

∴xC=xA=a，yB=yA=b=．                                                                       

∵點B、C在雙曲線y=上，                                                                   

∴xB==，yC=．                                                                           

∴點B的座標為（，），點C的座標為（a，）．                                   

∴AB=a﹣=，AC=﹣=．                                                             

∴S△ABC=ABAC                                                                               

=××                                                                                          

=．                                                                                                  

∴在點A運動過程中，△ABC的面積不變，始終等於．                               

（3）①AC為平行四邊形的一邊，                                                           

Ⅰ．當點B在點Q的右邊時，如圖3，                                                    

∵四邊形ACBD是平行四邊形，                                                              

∴AC∥BD，AC=BD．                                                                        

∴xD=xB=．                                                                                     

∴yD=2xD=．                                                                                   

∴DB=﹣．                                                                                    

∵AC=﹣=，                                                                               

∴=﹣．                                                                                      

解得：a=±2．                                                                               

經檢驗：a=±2是該方程的解．                                                            

∵a＞0，                                                                                            

∴a=2．                                                                                        

∴b==．                                                                                   

∴點A的座標為（2，）．                                                          

Ⅱ．當點B在點Q的左邊且點C在點Q的右邊時，如圖4，                           

∵四邊形ACDB是平行四邊形，                                                              

∴AC∥BD，AC=BD．                                                                        

∴xD=xB=．                                                                                     

∴yD=2xD=．                                                                                   

∴DB=﹣．                                                                                    

∵AC=，                                                                                          

∴=﹣，                                                                                      

解得：a=±2．                                                                                    

經檢驗：a=±2是該方程的解．                                                                 

∵a＞0，                                                                                            

∴a=2．                                                                                             

∴b==4．                                                                                         

∴點A的座標為（2，4）．                                                                     

②AC為平行四邊形的對角線，                                                                

此時點B、點C都在點Q的左邊，如圖5，                                              

∵四邊形ABCD是平行四邊形，                                                              

∴AB∥CD，AB=CD．                                                                        

∴yD=yC=．                                                                                     

∴xD==．                                                                                    

∴CD=﹣a．                                                                                    

∵AB=a﹣=，                                                                              

∴=﹣a．                                                                                     

解得：a=±．                                                                               

經檢驗：a=±是該方程的解．                                                            

∵a＞0，                                                                                            

∴a=．                                                                                        

∴b==4．                                                                                    

∴點A的座標為（，4）．                                                          

綜上所述：當點A、B、C、D為頂點的四邊形為平行四邊形時，此時點A的座標為（2，）或（2，4）或（，4）．                                                                                            

知識點：反比例函式

題型：綜合題