摘要：081]解:.b＝-.c＝-3．···························································· 3分 .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．··························································································· 4分 ①当H在Q.B之间时. QH＝OH-OQ ＝(4-4t)-4t＝4-8t．································································ 5分 ②当H在O.Q之间时. QH＝OQ-OH ＝4t-(4-4t)＝8t-4．································································ 6分 综合①.②得QH＝|4-8t|,································································ 6分 (3)存在t的值.使以P.H.Q为顶点的三角形与△COQ相似．···················· 7分 ①当H在Q.B之间时.QH＝4-8t. 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＝-1.t2＝--1．················································· 8分 ②当H在O.Q之间时.QH＝8t-4． 若△QHP∽△COQ.则QH∶CO＝HP∶OQ.得＝. ∴t＝．···························································································· 9分 若△PHQ∽△COQ.则PH∶CO＝HQ∶OQ.得＝. 即t2-2t+1＝0． ∴t1＝t2＝1．·········································································· 10分 综上所述.存在的值.t1＝-1.t2＝.t3＝．························ 10分 附加题:解:(1)8,···································································································· 5分 (2)2．·································································································· 10分

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