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1、二次函数和等腰三角形:(2008重庆)已知:如图,抛物线)0(22acaxaxy与y轴交于点C(0,4),与x轴交于点A、B,点A的坐标为(4,0)。(1)求该抛物线的解析式;(2)点Q是线段AB上的动点,过点Q作QE∥AC,交BC于点E,连接CQ。当△CQE的面积最大时,求点Q的坐标;(3)若平行于x轴的动直线l与该抛物线交于点P,与直线AC交于点F,点D的坐标为(2,0)。问:是否存在这样的直线l,使得△ODF是等腰三角形?若存在,请求出点P的坐标;若不存在,请说明理由。.解:(1)由题意,得01684aacc,.······················································(1分)解得124ac,.·························································································(2分)所求抛物线的解析式为:2142yxx.·············································(3分)(2)设点Q的坐标为(0)m,,过点E作EGx轴于点G.由21402xx,得12x,24x.点B的坐标为(20),.··········································································(4分)6AB,2BQm.QEAC∥,BQEBAC△∽△.EGBQCOBA,即246EGm.243mEG.··········(5分)CQECBQEBQSSS△△△YXCADQBO28题图1122BQCOBQEG124(2)423mm2128333mm····················(6分)21(1)33m.又24m≤≤,当1m时,CQES△有最大值3,此时(10)Q,.············································(7分)(3)存在.在ODF△中.(ⅰ)若DODF,(40)(20)AD,,,,2ADODDF.又在RtAOC△中,4OAOC,45OAC.45DFAOAC.90ADF.此时,点F的坐标为(22),.由21422xx,得115x,215x.此时,点P的坐标为:(152)P,或(152)P,.·······································(8分)(ⅱ)若FOFD,过点F作FMx轴于点M,由等腰三角形的性质得:112OMOD,3AM,在等腰直角AMF△中,3MFAM.(13)F,.由21432xx,得113x,213x.此时,点P的坐标为:(133)P,或(133)P,.········································(9分)(ⅲ)若ODOF,4OAOC,且9042AOCAC,,点O到AC的距离为22,而222OFOD,此时,不存在这样的直线l,使得ODF△是等腰三角形.······························(10分)综上所述,存在这样的直线l,使得ODF△是等腰三角形.所求点P的坐标为:(152)P,或(152)P,或(133)P,或(133)P,2.二次函数和矩形、等腰三角形:如图19-1,OABC是一张放在平面直角坐标系中的矩形纸片,O为原点,点A在x轴的正半轴上,点C在y轴的正半轴上,5OA,4OC.(1)在OC边上取一点D,将纸片沿AD翻折,使点O落在BC边上的点E处,求DE,两点的坐标;(2)如图19-2,若AE上有一动点P(不与AE,重合)自A点沿AE方向向E点匀速运动,运动的速度为每秒1个单位长度,设运动的时间为t秒(05t),过P点作ED的平行线交AD于点M,过点M作AE的平行线交DE于点N.求四边形PMNE的面积S与时间t之间的函数关系式;当t取何值时,S有最大值?最大值是多少?(3)在(2)的条件下,当t为何值时,以AME,,为顶点的三角形为等腰三角形,并求出相应的时刻点M的坐标.解:(1)依题意可知,折痕AD是四边形OAED的对称轴,在RtABE△中,5AEAO,4AB.2222543BEAEAB.2CE.E点坐标为(2,4).··············································································2分在RtDCE△中,222DCCEDE,又DEOD.222(4)2ODOD.解得:52CD.D点坐标为502,·················································································3分(2)如图①PMED∥,APMAED△∽△.PMAPEDAE,又知APt,52ED,5AE5522ttPM,又5PEt.而显然四边形PMNE为矩形.215(5)222PMNEtSPMPEttt矩形·············································5分21525228PMNESt四边形,又5052当52t时,PMNES矩形有最大值258.·························································6分(3)(i)若以AE为等腰三角形的底,则MEMA(如图①)在RtAED△中,MEMA,PMAE,P为AE的中点,1522tAPAE.yxBCOADE图5-1yxBCOADE图5-2PMNyxBCOADE图①PMNF又PMED∥,M为AD的中点.过点M作MFOA,垂足为F,则MF是OAD△的中位线,1524MFOD,1522OFOA,当52t时,5052,AME△为等腰三角形.此时M点坐标为5524,.·8分(ii)若以AE为等腰三角形的腰,则5AMAE(如图②)在RtAOD△中,2222555522ADODAO.过点M作MFOA,垂足为F.PMED∥,APMAED△∽△.APAMAEAD.5525552AMAEtAPAD,152PMt.5MFMP,525OFOAAFOAAP,当25t时,(0255),此时M点坐标为(5255),.················11分综合(i)(ii)可知,52t或25t时,以AME,,为顶点的三角形为等腰三角形,相应M点的坐标为5524,或(5255),.3、二次函数和梯形:(2009临沂)如图,已知抛物线与x轴交于A(-1,0)、B(3,0)两点,与y轴交于点C(0,3)。⑴求抛物线的解析式;⑵设抛物线的顶点为D,在其对称轴的右侧的抛物线上是否存在点P,使得△PDC是等腰三角形?若存在,求出符合条件的点P的坐标;若不存在,请说明理由;⑶若点M是抛物线上一点,以B、C、D、M为顶点的四边形是直角梯形,试求出点M的坐标。⑴∵抛物线与y轴交于点C(0,3),yxBCOADE图②PMNF第26题图xyAMPDOBC∴设抛物线解析式为)0(32abxaxy………………………………1分根据题意,得,0339,03baba,解得.2,1ba∴抛物线的解析式为322xxy………………………………………2分⑵存在。…………………………………………………………………………3分由322xxy得,D点坐标为(1,4),对称轴为x=1。…………4分①若以CD为底边,则PD=PC,设P点坐标为(x,y),根据勾股定理,得2222)4()1()3(yxyx,即y=4-x。…………………………5分又P点(x,y)在抛物线上,∴3242xxx,即0132xx…………6分解得253x,1253,应舍去。∴253x。……………………7分∴2554xy,即点P坐标为255,253。……………………8分②若以CD为一腰,因为点P在对称轴右侧的抛物线上,由抛物线对称性知,点P与点C关于直线x=1对称,此时点P坐标为(2,3)。∴符合条件的点P坐标为255,253或(2,3)。……………………9分⑶由B(3,0),C(0,3),D(1,4),根据勾股定理,得CB=23,CD=2,BD=52,………………………………………………10分∴20222BDCDCB,∴∠BCD=90°,………………………………………………………………………11分设对称轴交x轴于点E,过C作CM⊥DE,交抛物线于点M,垂足为F,在Rt△DCF中,∵CF=DF=1,∴∠CDF=45°,由抛物线对称性可知,∠CDM=2×45°=90°,点坐标M为(2,3),∴DM∥BC,∴四边形BCDM为直角梯形,………………………………………………………12分由∠BCD=90°及题意可知,以BC为一底时,顶点M在抛物线上的直角梯形只有上述一种情况;以CD为一底或以BD为一底,且顶点M在抛物线上的直角梯形均不存在。综上所述,符合条件的点M的坐标为(2,ExyAMPDOBCF3)。…………………………………13分4、动态、二次函数、相似(2009年济南)如图,在梯形ABCD中,354245ADBCADDCABB∥,,,,∠.动点M从B点出发沿线段BC以每秒2个单位长度的速度向终点C运动;动点N同时从C点出发沿线段CD以每秒1个单位长度的速度向终点D运动.设运动的时间为t秒.(1)求BC的长.(2)当MNAB∥时,求t的值.(3)试探究:t为何值时,MNC△为等腰三角形.【关键词】动态问题、二次函数、相似三角形【答案】(1)如图①,过A、D分别作AKBC于K,DHBC于H,则四边形ADHK是矩形∴3KHAD.在RtABK△中,2sin454242AKAB.,2cos454242BKAB,在RtCDH△中,由勾股定理得,22543HC∴43310BCBKKHHC,DCBMN(2)如图②,过D作DGAB∥交BC于G点,则四边形ADGB是平行四边形,∵MNAB∥∴MNDG∥,∴3BGAD,∴1037GC,由题意知,当M、N运动到t秒时,102CNtCMt,.∵DGMN∥,∴NMCDGC∠∠,又CC∠∠,∴MNCGDC△∽△∴CNCMCDCG,即10257tt,解得,5017t,(3)分三种情况讨论:①当NCMC时,如图③,即102tt∴103t,(图①)ADCBKH图②ADCBGMNADCBMN图③)图④)ADCBMNHE②当MNNC时,如图④,过N作NEMC于E解法一:由等腰三角形三线合一性质得11102522ECMCtt,在RtCEN△中,5cosECtcNCt,又在RtDHC△中,3cos5CHcCD,∴535tt,解得258t,解法二:∵90CCDHCNEC∠∠,,∴NECDHC△∽△,∴NCECDCHC即553tt,∴258t,③当MNMC时,如图⑤,过M作MFCN于F点.1122FCNCt解法一:(方法同②中解法一)132cos1025tFCCMCt解得6017t解法二:∵90C
本文标题:二次函数拔高综合题全集(含答案)
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