Question

如图，已知抛物线y＝x²＋bx＋c与坐标轴交于A、B、C三点， A点的坐标为

（－1，0），过点C的直线y＝x－3与x轴交于点Q，点P是线段BC上的一个动点，过P作PH⊥OB于点H．若PB＝5t，且0＜t＜1.

1.（1）填空：点C的坐标是_       _，b＝_      _；

2.（2）求线段QH的长（用含t的式子表示）；

3.（3）依点P的变化，是否存在t的值，使以P、H、Q为顶点的三角形与△COQ相似？若存在，求出所有t的值；若不存在，说明理由.

Answer

1.（1）（0，－3），b＝－.

2.（2）由（1），得y＝x2－x－3，它与x轴交于A，B两点，得B（4，0）．

∴OB＝4，又∵OC＝3，∴BC＝5．

由题意，得△BHP∽△BOC，

∵OC∶OB∶BC＝3∶4∶5，

∴HP∶HB∶BP＝3∶4∶5，

∵PB＝5t，∴HB＝4t，HP＝3t．

∴OH＝OB－HB＝4－4t．

由y＝x－3与x轴交于点Q，得Q（4t，0）．

∴OQ＝4t．

①当H在Q、B之间时，

QH＝OH－OQ

＝（4－4t）－4t＝4－8t．································································ 3分

②当H在O、Q之间时，

QH＝OQ－OH

＝4t－（4－4t）＝8t－4．································································ 4分

综合①，②得QH＝｜4－8t｜；

3.（3）存在t的值，使以P、H、Q为顶点的三角形与△COQ相似．

①当H在Q、B之间时，QH＝4－8t，

若△QHP∽△COQ，则QH∶CO＝HP∶OQ，得＝，

∴t＝．···························································································· 5分

若△PHQ∽△COQ，则PH∶CO＝HQ∶OQ，得＝，

即t2＋2t－1＝0．

∴t1＝－1，t2＝－－1（舍去）．················································ 6分

②当H在O、Q之间时，QH＝8t－4．

若△QHP∽△COQ，则QH∶CO＝HP∶OQ，得＝，

∴t＝．···························································································· 7分

若△PHQ∽△COQ，则PH∶CO＝HQ∶OQ，得＝，

即t2－2t＋1＝0．

∴t1＝t2＝1（舍去）．············································································ 8分

综上所述，存在的值，t1＝－1，t2＝，t3＝．

【解析】略