上海市2020届初三数学一模提升题汇编第24题(二次函数综合)

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【2020长宁金山一模】24.(本题满分12分,每小题4分)如图,在平面直角坐标系xOy中,抛物线nmxxy231经过点)1,6(B、)0,5(C,且与y轴交于点A.(1)求抛物线的表达式及点A的坐标;(2)点P是y轴右侧抛物线上的一点,过点P作OAPQ,交线段OA的延长线于点Q,如果45PAB,求证:PQAΔ∽ACBΔ;(3)若点F是线段AB(不包含端点)上的一点,且点F关于AC的对称点F恰好在上述抛物线上,求FF的长.(长宁金山)24.(本题满分12分,每小题4分)解:(1)∵抛物线nmxxy231过点)1,6(B、)0,5(C∴055311663122nmnm∴538nm(2分)∴538312xxy(1分)令0x得5y,∴点A的坐标为)5,0((1分)(2)∵)5,0(A,)1,6(B,)0,5(C∴25AC,2BC,132AB∴222BCACAB∴90ACB又∵OAPQ∴90PQA∴ACBPQA(1分)∵)5,0(A,)0,5(C∴OCOA,∵90AOC∴45OCAOAC(1分)∵180CAOBACPABQAP,45PAB第24题图yxCABO∴90BACQAP∵90BACABC∴ABCQAP(1分)∴PQAΔ∽ACBΔ(1分)(3)设点B是点B关于直线AC的对称点,则2BCCB,90ACBBAC过点B作xBG轴,垂足为点G∵90OCACOB,45OCA,∴45COB∴1GCGB∴),(1-4B(1分)∵点F同时在线段BA与抛物线上,∴设)53831,(F2xxx分别过点F,B作轴yHF,轴yHB,垂足分别为H、H,则HBH//F∴HAAHHBHFBAFA即6313842xxx∴27x(1分)又∵ACFF,ACBB∴B//BFF∴BBFFBAFA∴87427HBHFBBFF(1分)∵222BCBB∴8722FF∴247FF(1分)【2020杨浦一模】已知在平面直角坐标系xOy中,抛物线ymx22mx4m0与x轴交于点A、B(点A在点B的左侧),且AB=6.(1)求这条抛物线的对称轴及表达式;(2)在y轴上取点E(0,2),点F为第一象限内抛物线上一点,联结BF、EF,如果S四边形OEFB10,求点F的坐标;(3)在第(2)小题的条件下,点F在抛物线对称轴右侧,点P在x轴上且在点B左侧,如果直线PF与y轴的夹角等于∠EBF,求点P的坐标.(杨浦)24.解:(1)抛物线对称轴212mxm-=-=.....................................................(1分)∵AB=6,∴抛物线与x轴的交点A为(20),-,B(40),...................................................(1分)∴4440mm++=(或16840mm-+=)..................................................................(1分)∴12m=-.∴抛物线的表达式为2142yxx=-++.......................................................(1分)(2)设点F21(4)2xxx,-++........................................................................................(1分)∵点E02(,),点B4(,0),∴OE=2,OB=4.∵=+10OEFOBFOEFBSSS四边形,∴211124(4)10222xxx......................(1分)∴12x或,∴点F912(,)、24(,)..................................................................................(2分)(3)∵=+10OBEBEFOEFBSSS四边形,又1142422OBESOBOE,∴6BEFS.过F作FHBE,垂足为点H.∵162BEFSBEFH,又222425BE,∴655FH................................(1分)又22(24)(40)25BF,∴855BH.∴在RtBFH中,tan∠EBF=65358455FHBH....................................................................(1分)设直线PF与y轴的交点为M,则∠PMO=∠EBF,过F作FGx轴,垂足为点G.∵FG//y轴,∴∠PMO=∠PFG.∴tan∠PFG=tan∠EBF.....................................................(1分)∴tan∠PFG=34PGFG.又FG=4,∴PG=3.∴点P的坐标10(-,)............................................................................................................(1分)【2020徐汇一模】24.(本题满分12分)如图,将抛物线4342xy平移后,新抛物线经过原抛物线的顶点C,新抛物线与x轴正半轴交于点B,联结BC,4tanB,设新抛物线与x轴的另一交点是A,新抛物线的顶点是D.(1)求点D的坐标;(2)设点E在新抛物线上,联结AC、DC,如果CE平分DCA,求点E的坐标;(3)在(2)的条件下,将抛物线4342xy沿x轴左右平移,点C的对应点为F,当DEF和ABC相似时,请直接写出平移后所得抛物线的表达式.(第24题图)ADCBOxy(徐汇)24.解:(1)由题意,设新抛物线的表达式为4342bxxy.∵抛物线4342xy的顶点为C,∴)4,0(C,4OC;在BOCRt中,90BOC,∴4tanOBOCB,得1OB;∴)0,1(B;由题意,得0434b,解得38b;∴新抛物线的表达式为438342xxy;∴)316,1(D.(2)由题意,可得)0,3(A;过点D作OCDM,垂足为M.∴)316,0(M;∴4,3,34,1COAOCMDM;∴43COAOCMDM;又90AOCDMC,∴DMC∽AOC,∴ACODCM;∵CE平分DCA,∴ACEDCE;∴180)(2DCEDCM;∴AOCMCE90;∴AOCE//;∴点E与点C关于直线1x对称;∴)4,2(E.(3)有两种情况满足要求,平移后所得抛物线的表达式为:4)32(342xy或4)121(342xy【2020松江一模】24.(本题满分12分,第(1)小题3分,第(2)小题4分,第(3)小题5分)如图,已知抛物线y=﹣x2+bx+c经过点A(3,0),点B(0,3).点M(m,0)在线段OA上(与点A,O不重合),过点M作x轴的垂线与线段AB交于点P,与抛物线交于点Q,联结BQ.(1)求抛物线表达式;(2)联结OP,当∠BOP=∠PBQ时,求PQ的长度;(3)当△PBQ为等腰三角形时,求m的值.24.解:(1)∵抛物线y=﹣x2+bx+c经过点A(3,0),点B(0,3).∴3,930.cbc………………………………(1分)∴b=2,c=3………(1分)∴抛物线表达式为y=﹣x2+2x+3………(1分)(2)∵PM⊥x轴∴PM∥y轴∴∠OBP=∠BPQ∵∠BOP=∠PBQ∴△OBP∽△BPQ………………(1分)∴OBBPBPPQ∴2BPOBPQ………(1分)∴22(2)3(2+3+3)mmmm即222-39mmm解得95m(m=0舍去)………(1分)5425PQ………(1分)(3)当QP=QP时(第24题备用图)yxOBA(第24题图)yxOBAPMQ(第24题图)yxOBAPMQ点Q(2,3)此时m=2………(1分)当BQ=BP时,点Q(1,4)此时m=1………(2分)当PB=PQ时22233mmmm32m………(2分)【2020青浦一模】24.(本题满分12分,其中第(1)小题4分,第(2)小题5分,第(3)小题3分)如图,在平面直角坐标系xOy中,抛物线2yxbxc与x轴交于A、B两点,与y轴交于点C,对称轴为直线x=2,点A的坐标为(1,0).(1)求该抛物线的表达式及顶点坐标;(2)点P为抛物线上一点(不与点A重合),联结PC.当∠PCB=∠ACB时,求点P的坐标;(3)在(2)的条件下,将抛物线沿平行于y轴的方向向下平移,平移后的抛物线的顶点为点D,点P的对应点为点Q,当OD⊥DQ时,求抛物线平移的距离.24.解:(1)∵A的坐标为(1,0),对称轴为直线x=2,∴点B的坐标为(3,0)····(1分)将A(1,0)、B(3,0)代入2+yxbxc,得10930.,bcbc解得:43.,bc·························································(2分)所以,243yxx.yxCBAOyxCBAO当x=2时,2242+3=1y∴顶点坐标为(2,-1)·················································································(1分).(2)过点P作PN⊥x轴,垂足为点N.过点C作CM⊥PN,交NP的延长线于点M.∵∠CON=90°,∴四边形CONM为矩形.∴∠CMN=90°,CO=MN.∵243yxx,∴点C的坐标为(0,3)····················································(1分).∵B(3,0),∴OB=OC.∵∠COB=90°,∴∠OCB=∠BCM=45°,····················(1分).又∵∠ACB=∠PCB,∴∠OCB-∠ACB=∠BCM-∠PCB,即∠OCA=∠PCM.······(1分).∴tan∠OCA=tan∠PCM.∴13PMMC.设PM=a,则MC=3a,PN=3-a.∴P(3a,3-a).·······························································································(1分)将P(3a,3-a)代入243yxx,得231233aaa.解得111=9a,2=0a(舍).∴P(113,169).····················································(1分)(3)设抛物线平移的距离为m.得221yxm,∴D的坐标为(2,1m).·····················································································(1分)过点D作直线EF∥x轴,交y轴于点E,交PQ的延长线于点F.∵∠OED=∠QFD=∠ODQ=90°,∴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