上海市2019届初三数学一模提升题汇编第24题(二次函数综合)(含2019上海中考试题答案)

1上海市2019届一模提升题汇编第24题（二次函数综合）含2019上海中考试题中考【2019届一模徐汇】24．（本题满分12分，第（1）小题4分，第（2）小题4分，第（3）小题4分）如图，在平面直角坐标系xoy中，顶点为M的抛物线C1:2(0)yaxbxa经过点A和x轴上的点B，AO=OB=2，120AOBo．（1）求该抛物线的表达式；（2）联结AM，求AOMSV；（3）将抛物线C1向上平移得到抛物线C2，抛物线C2与x轴分别交于点E、F（点E在点F的左侧），如果△MBF与△AOM相似，求所有符合条件的抛物线C2的表达式．【24．解：（1）过A作AH⊥x轴，垂足为H，∵OB=2，∴B（2,0）………………………………（1分）∵120AOB∴60,30AOHHAO．∵OA=2，∴112OHOA．∵222RtAHOOHAHOAV在中，，∴22213AH．∴(1,3)A……………………………………（1分）（第24题图）2∵抛物线21:CyaxbxAB经过点、，∴可得：342033233aababb解得：………………………………………………（1分）∴这条抛物线的表达式为232333yxx…………………………………………（1分）（2）过M作MG⊥x轴，垂足为G，∵232333yxx∴顶点M是31,3，得33MG……………………………………………………（1分）∵(1,3)A，M31,3．∴得：直线AM为23333yx…………………………………………………（1分）∴直线AM与x轴的交点N为1,02……………………………………………………（1分）∴1122AOMSONMGONAH1131132232233…………………………………………………………………………（1分）（3）∵）33，1（M、)0,2(B，∴33MGRtBGMMBGBG在中，tan=，∴MBG=30．∴MBF150．由抛物线的轴对称性得：MO=MB，3∴MBOMOB=150．∵OB=120A，∴OM=150A∴OM=MBFA．∴BMBFOAOM或BFBMOAOM相似时，有：AOM与MBF当即332BF2332或BF3322332，∴32BF或2BF．∴）0，38）或（0，4（F………………………………………………（2分）设向上平移后的抛物线kxxy33233:为C22，当）0，4（F时，338k，∴抛物线33833233:为C22xxy…（1分）当）0，38（F时，27316k，抛物线22323163:3327Cyxx……．（1分）】【2019届一模浦东】24.（本题满分12分，其中每小题各4分）已知：如图9，在平面直角坐标系xOy中，直线12yxb与x轴相交于点A，与y轴相交于点B.抛物线244yaxax经过点A和点B，并与x轴相交于另一点C，对称轴与x轴相交于点D.（1）求抛物线的表达式;（2）求证:△BOD∽△AOB;（3）如果点P在线段AB上，且∠BCP=∠DBO，求点P的坐标.（图9）xBOAy4【24、（1）211482yxx；（2）证明略；（3）1612,55】【2019届一模杨浦】24．（本题满分12分，每小题各4分）在平面直角坐标系xOy中，抛物线2(0)yaxbxca=++?与y轴交于点C（0,2），它的顶点为D（1,m），且1tan3COD?.（1）求m的值及抛物线的表达式；（2）将此抛物线向上平移后与x轴正半轴交于点A，与y轴交于点B，且OA=OB.若点A是由原抛物线上的点E平移所得，求点E的坐标；（3）在（2）的条件下，点P是抛物线对称轴上的一点（位于x轴上方），且∠APB=45°.求P点的坐标.Oxy123412345-1-2-3-1-2-3（第24题图）5【24．解：（1）作DH⊥y轴，垂足为H，∵D（1,m）（0m），∴DH=m，HO=1.∵1tan3COD?，∴13OHDH=，∴m=3.·····················································（1分）∴抛物线2yaxbxc=++的顶点为D（1,3）.又∵抛物线2yaxbxc=++与y轴交于点C（0,2），∴3,1,22.abcbacì++=ïïïïï-=íïïïï=ïî（2分）∴1,2,2.abcì=-ïïïï=íïï=ïïî∴抛物线的表达式为222yxx=-++.······（1分）（2）∵将此抛物线向上平移，∴设平移后的抛物线表达式为222(0)yxxkk=-+++，.····························（1分）则它与y轴交点B（0,2+k）.∵平移后的抛物线与x轴正半轴交于点A，且OA=OB，∴A点的坐标为（2+k,0）..（1分）∴20(2)2(2)2kkk=-+++++.∴122,1kk=-=.∵0k，∴1k=.∴A（3,0），抛物线222yxx=-++向上平移了1个单位..······························（1分）∵点A由点E向上平移了1个单位所得，∴E（3,-1）..···································（1分）（3）由（2）得A（3,0），B（0,3），∴32AB=.∵点P是抛物线对称轴上的一点（位于x轴上方），且∠APB=45°,原顶点D（1,3）,∴设P（1,y），设对称轴与AB的交点为M，与x轴的交点为H，则H（1,0）.∵A（3,0），B（0,3），∴∠OAB=45°,∴∠AMH=45°.∴M（1,2）.∴2BM=.∵∠BMP=∠AMH,∴∠BMP=45°.∵∠APB=45°,∴∠BMP=∠APB.∵∠B=∠B，∴△BMP∽△BPA.··································································（2分）BAPxyOMH6∴BPBABMBP=.∴23226BPBABM=??∴221(3)6BPy=+-=.∴123535yy，=+=-（舍）..····························（1分）∴(1,35)P+..·····················································································（1分）】【2019届一模普陀】24．（本题满分12分）如图10，在平面直角坐标系中，抛物线23yaxbx(0)a与x轴交于点A1,0和点B，且3OBOA，与y轴交于点C，此抛物线顶点为点D．（1）求抛物线的表达式及点D的坐标；（2）如果点E是y轴上的一点（点E与点C不重合），当BEDE时，求点E的坐标；（3）如果点F是抛物线上的一点，且，求点F的坐标．135FBDxOy图10CBAOyx7【24．解：（1）∵抛物线与x轴交于点A1,0和点，且3OBOA，∴点的坐标是3,0．···········································································（1分）解法一：由抛物线23yaxbx经过点1,0和3,0．得03,0933.abab解得1,2.ab······························································（1分）∴抛物线的表达式是223yxx．······················································（1分）点D的坐标是1,4．·············································································（1分）解法二：由抛物线23yaxbx经过点1,0和3,0．可设抛物线的表达式为(1)(3)yaxx，由抛物线与y轴的交点C的坐标是0,3，得3(01)(03)a，解得1a．······························································（1分）∴抛物线的表达式是223yxx．························································（1分）点D的坐标是1,4．·············································································（1分）（2）过点D作DHOC，H为垂足．∴90DHO．∴90DEHEDH．∵BEDE，∴90DEHBEO．∴BEOEDH．又∵BOEEHD，∴△BOE∽△EHD．·········································（1分）∴BOOEEHHD．∵点D的坐标是1,4，∴1DH，4OH．BB8∵点的坐标是3,0，∴3OB．∴341OEOE．··············································································（1分）∴1OE或3OE．················································································（1分）∵点E与点C不重合，∴1OE．∴点E的坐标是0,1．···········································································（1分）（3）过点F作FGx轴，G为垂足．作45DBM，由第（2）题可得，点M与点E重合．∵1OE，1DH，∴OEDH．可得△BOE≌△EHD．∴BEED．∵90BED，∴45DBE．∵135FBD，∴90FBE．················································································（1分）∴OBEGFB．∴在Rt△BOE中，90BOE，∴cot3OBE∴cot3GFB．··········（1分）∴3FGBG．设点F点的坐标为2,23mmm．∴223FGmm,3BGm．∴2233(3)mmm．···································································（1分）解得3m，4m．∵3m不合题意舍去，∴4m．点F的坐标是4,21．··········································································（1分）】【2019届一模奉贤】24．（本题满分12分，每小题满分6分）B9如图10，在平面直角坐标系中，直线AB与抛物线2yaxbx=+交于点A(6，0)和点B(1，－5)．（1）求这条抛物线的表达式和直线AB的表达式；（2）如果点C在直线AB上，且∠BOC的正切值是32，求点C的坐标．【24．解：（1）由题意得，抛物线2yaxbx=+经过点A(6，0)和点B(1，－5)，代入得3660,5.ababì+=ïïíï+=-ïî解得1,6.abì=ïïíï=-ïî∴抛物线