2019-2020學(xué)年江蘇省蘇州市常熟市九年級(jí)(上)期末數(shù)學(xué)試卷

第第22頁（共19頁）答：BC的長為4匚5.【解答】證明：(l)TZBAC=90°，點(diǎn)G是BC的中點(diǎn)，:.AG=BG=GC,AZABG=ZBAG,又..勺。丄AG,AZBAG+ZDAF=ZADF+ZDAF=90°,AZADB=ZBAG,???怔二AB,AZADB=ZAEB,\ZABE=ZAEB,:.AB=AE,VQO是AABD的外接圓，且ZBAD=90°,；?BD是直徑，DM是OO切線，:,DM丄BD,且BD丄AG,.??DM〃AG,型…'CG??塑=丄.而_H，.亜一?——,CACG4設(shè)CD=3k,AC=4k,?.AD=k,VZBDA=ZABC,ZBAD=ZCAB,.?.△ABDsAacb,..-,ABAC.??AB2=AD?AC=4k2,?.AB=2k,?\tanZABC==2；ABFAEF.DF=1,tanZABC=tanZADF=tanZBAF==^-=2,:.AF=2,BF=4,...AB=匚拇z十程十16=2T5,.??AC=4l5,.°.BC='[凰2十應(yīng)c?=二20十20=10,:.BG=5,【解答】解：(1)T拋物線y=ax2+bx+c(aMO)的頂點(diǎn)為A(-2,0),???設(shè)拋物線的解析式為y=a(x+2)2,將點(diǎn)B(-5,9)代入y=a(x+2)2中，得，9=a(-5+2)2,?a=1,?拋物線的解析式為y=(x+2)2=x2+4x+4；(2)①如圖①，由(1)知，拋物線的解析式為y=x2+4x+4,C(0,4),?:B(-5,9),??直線BC的解析式為y=-x+4,過點(diǎn)A作AH〃y軸，交直線BC于H,VA(-2,0),H(-2,6),.??Smbc=￡hX(xc-xb)=丄X6X5=15,TS△PAB=^S△ABC，?SaP4B=TX15=3，VA(-2,0),B(-5,9),?直線AB的解析式為y=-3x-6設(shè)點(diǎn)P(p,p2+4p+4),G(p,-3p-6),?.SaP4B—[-3p-6-(p2+4p+4)]X(-2+5)=3,?p=-3或p=-4,?P(-3,1)或(-4,4)；②如圖②，?.?BD丄x軸，且B(-5,9),???D(-5,0),設(shè)直線BN的解析式為y=k(x+5)+9①,令y=0,貝9k(x+5)+9=0,TOC\o"1-5"\h\z.陸+9=9..x=-=-5—kk9:,N(-5,0),QQ.°.DN=-5-上+5=,kk???點(diǎn)A(-2,0),???設(shè)直線AM的解析式為y=k'(x+2)②,當(dāng)x=5時(shí),y=-3k',AM(-5,-3k'),:.DM=-3k,TOC\o"1-5"\h\z.??P(-2-2X，,-3kX■),k-kyk-k;???點(diǎn)P在拋物線y=(x+2)2上,???(-2-3X，+2)2=-3k'X，,k-krk-ky.3k+9_,j/.k=k'-3,91I.DN(DM+DB)=-丄(-3k'+9)=27X上(k'-3)=27X土Xk=27；kkk即：DN(DM+DB)為定值27.

?CD=2?,AD3V/r/