Question

如图，直线y=x+2与抛物线y=ax²+bx+6（a≠0）相交于A（，）和B（4，m），点P是线段AB上异于A、B的动点，过点P作PC⊥x轴于点D，交抛物线于点C．                                             

（1）求抛物线的解析式；                                                                       

（2）是否存在这样的P点，使线段PC的长有最大值？若存在，求出这个最大值；若不存在，请说明理由；                

（3）求△PAC为直角三角形时点P的坐标．                                           

                                                              

　                                                                                                      

                                                                                                          

Answer 1

【考点】二次函数综合题．                                                                     

【专题】几何综合题；压轴题．                                                              

【分析】（1）已知B（4，m）在直线y=x+2上，可求得m的值，抛物线图象上的A、B两点坐标，可将其代入抛物线的解析式中，通过联立方程组即可求得待定系数的值．                                         

（2）要弄清PC的长，实际是直线AB与抛物线函数值的差．可设出P点横坐标，根据直线AB和抛物线的解析式表示出P、C的纵坐标，进而得到关于PC与P点横坐标的函数关系式，根据函数的性质即可求出PC的最大值．                    

（3）当△PAC为直角三角形时，根据直角顶点的不同，有三种情形，需要分类讨论，分别求解．                  

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

∴m=4+2=6，                                                                                     

∴B（4，6），                                                                                   

∵A（，）、B（4，6）在抛物线y=ax²+bx+6上，                                     

∴，解得，                                              

∴抛物线的解析式为y=2x²﹣8x+6．                                                        

                                                                                                          

（2）设动点P的坐标为（n，n+2），则C点的坐标为（n，2n²﹣8n+6），                  

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

=﹣2n²+9n﹣4，                                                                                 

=﹣2（n﹣）²+，                                                                        

∵PC＞0，                                                                                         

∴当n=时，线段PC最大且为．                                                       

                                                                                                          

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

i）若点P为直角顶点，则∠APC=90°．                                                   

由题意易知，PC∥y轴，∠APC=45°，因此这种情形不存在；                        

ii）若点A为直角顶点，则∠PAC=90°．                                                  

如答图3﹣1，过点A（，）作AN⊥x轴于点N，则ON=，AN=．                

过点A作AM⊥直线AB，交x轴于点M，则由题意易知，△AMN为等腰直角三角形，              

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

∴M（3，0）．                                                                                  

设直线AM的解析式为：y=kx+b，                                                          

则：，解得，                                                             

∴直线AM的解析式为：y=﹣x+3  ①                                                     

又抛物线的解析式为：y=2x²﹣8x+6 ②                                                    

联立①②式，解得：x=3或x=（与点A重合，舍去）                                   

∴C（3，0），即点C、M点重合．                                                         

当x=3时，y=x+2=5，                                                                        

∴P₁（3，5）；                                                                                 

              

iii）若点C为直角顶点，则∠ACP=90°．                                                  

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

∴抛物线的对称轴为直线x=2．                                                               

如答图3﹣2，作点A（，）关于对称轴x=2的对称点C，                          

则点C在抛物线上，且C（，）．                                                      

当x=时，y=x+2=．                                                                          

∴P₂（，）．                                                                              

∵点P₁（3，5）、P₂（，）均在线段AB上，                                          

∴综上所述，△PAC为直角三角形时，点P的坐标为（3，5）或（，）．                    

【点评】此题主要考查了二次函数解析式的确定、二次函数最值的应用以及直角三角形的判定、函数图象交点坐标的求法等知识．