Question

如圖，直線y=x+2與拋物線y=ax²+bx+6（a≠0）相交于A（，）和B（4，m），點P是線段AB上異于A、B的動點，過點P作PC⊥x軸于點D，交拋物線于點C．                                             

（1）求拋物線的解析式；                                                                       

（2）是否存在這樣的P點，使線段PC的長有最大值？若存在，求出這個最大值；若不存在，請說明理由；                

（3）求△PAC為直角三角形時點P的坐標．                                           

                                                              

　                                                                                                      

                                                                                                          

Answer 1

【考點】二次函數(shù)綜合題．                                                                     

【專題】幾何綜合題；壓軸題．                                                              

【分析】（1）已知B（4，m）在直線y=x+2上，可求得m的值，拋物線圖象上的A、B兩點坐標，可將其代入拋物線的解析式中，通過聯(lián)立方程組即可求得待定系數(shù)的值．                                         

（2）要弄清PC的長，實際是直線AB與拋物線函數(shù)值的差．可設(shè)出P點橫坐標，根據(jù)直線AB和拋物線的解析式表示出P、C的縱坐標，進而得到關(guān)于PC與P點橫坐標的函數(shù)關(guān)系式，根據(jù)函數(shù)的性質(zhì)即可求出PC的最大值．                    

（3）當△PAC為直角三角形時，根據(jù)直角頂點的不同，有三種情形，需要分類討論，分別求解．                  

【解答】解：（1）∵B（4，m）在直線y=x+2上，                                        

∴m=4+2=6，                                                                                     

∴B（4，6），                                                                                   

∵A（，）、B（4，6）在拋物線y=ax²+bx+6上，                                     

∴，解得，                                              

∴拋物線的解析式為y=2x²﹣8x+6．                                                        

                                                                                                          

（2）設(shè)動點P的坐標為（n，n+2），則C點的坐標為（n，2n²﹣8n+6），                  

∴PC=（n+2）﹣（2n²﹣8n+6），                                                           

=﹣2n²+9n﹣4，                                                                                 

=﹣2（n﹣）²+，                                                                        

∵PC＞0，                                                                                         

∴當n=時，線段PC最大且為．                                                       

                                                                                                          

（3）∵△PAC為直角三角形，                                                                

i）若點P為直角頂點，則∠APC=90°．                                                   

由題意易知，PC∥y軸，∠APC=45°，因此這種情形不存在；                        

ii）若點A為直角頂點，則∠PAC=90°．                                                  

如答圖3﹣1，過點A（，）作AN⊥x軸于點N，則ON=，AN=．                

過點A作AM⊥直線AB，交x軸于點M，則由題意易知，△AMN為等腰直角三角形，              

∴MN=AN=，∴OM=ON+MN=+=3，                                                

∴M（3，0）．                                                                                  

設(shè)直線AM的解析式為：y=kx+b，                                                          

則：，解得，                                                             

∴直線AM的解析式為：y=﹣x+3  ①                                                     

又拋物線的解析式為：y=2x²﹣8x+6 ②                                                    

聯(lián)立①②式，解得：x=3或x=（與點A重合，舍去）                                   

∴C（3，0），即點C、M點重合．                                                         

當x=3時，y=x+2=5，                                                                        

∴P₁（3，5）；                                                                                 

              

iii）若點C為直角頂點，則∠ACP=90°．                                                  

∵y=2x²﹣8x+6=2（x﹣2）²﹣2，                                                             

∴拋物線的對稱軸為直線x=2．                                                               

如答圖3﹣2，作點A（，）關(guān)于對稱軸x=2的對稱點C，                          

則點C在拋物線上，且C（，）．                                                      

當x=時，y=x+2=．                                                                          

∴P₂（，）．                                                                              

∵點P₁（3，5）、P₂（，）均在線段AB上，                                          

∴綜上所述，△PAC為直角三角形時，點P的坐標為（3，5）或（，）．                    

【點評】此題主要考查了二次函數(shù)解析式的確定、二次函數(shù)最值的應(yīng)用以及直角三角形的判定、函數(shù)圖象交點坐標的求法等知識．