初中数学动点问题及答案

•例1（2009年兰州）如图①，正方形ABCD中，点A、B的坐标分别为（0，10），（8，4），点C在第一象限．动点P在正方形ABCD的边上，从点A出发沿A→B→C→D匀速运动，同时动点Q以相同速度在x轴正半轴上运动，当P点到达D点时，两点同时停止运动，设运动的时间为t秒．(1)当P点在边AB上运动时，点Q的横坐标（长度单位）关于运动时间t（秒）的函数图象如图②所示，请写出点Q开始运动时的坐标及点P运动速度；(2)求正方形边长及顶点C的坐标；•(3)在（1）中当t为何值时，△OPQ的面积最大，并求此时P点的坐标；•(4)如果点P、Q保持原速度不变，当点P沿A→B→C→D匀速运动时，OP与PQ能否相等，若能，写出所有符合条件的t的值；若不能，请说明理由．注意：第（4）问按点P分别在AB、BC、CD边上分类讨论；求t值时，灵活运用等腰三角形“三线合一”。•例2（2009年齐齐哈尔市）直线364yx与坐标轴分别交于A、B两点，动点P、Q同时从O点出发，同时到达A点，运动停止．点Q沿线段OA运动，速度为每秒1个单位长度，点P沿路线O→B→A运动．•（1）直接写出A、B两点的坐标；•（2）设点Q的运动时间为t秒，△OPQ的面积为S，求出S与t之间的函数关系式；•（3）当S=48/5时，求出点P的坐标，并直接写O、P、Q出以点为顶点的平行四边形的第四个顶点M的坐标．•例3(2009年哈尔滨）如图1，在平面直角坐标系中，点O是坐标原点，四边形ABCO是菱形，点A的坐标为（-3，4），点C在x轴的正半轴上，直线AC交y轴于点M，AB边交y轴于点H．（1）求直线AC的解析式；（2）连接BM，如图2，动点P从点A出发，沿折线ABC方向以2个单位／秒的速度向终点C匀速运动，设△PMB的面积为S（S不为0），点P的运动时间为t秒，求S与t之间的函数关系式（要求写出自变量t的取值范围）；（3）在（2）的条件下，当t为何值时，∠MPB与∠BCO互为余角，并求此时直线OP与直线AC所夹锐角的正切值．图（1）•例4（2009年湖北十堰市）如图①，已知抛物线32bxaxy（a≠0）与轴交于点A(1，0)和点B(－3，0)，与y轴交于点C．•(1)求抛物线的解析式；•(2)设抛物线的对称轴与轴交于点M，问在对称轴上是否存在点P，使△CMP为等腰三角形？若存在，请直接写出所有符合条件的点P的坐标；若不存在，请说明理由．(3)如图②，若点E为第二象限抛物线上一动点，连接BE、CE，求四边形BOCE面积的最大值，并求此时E点的坐标．例1解：（1）Q（1，0）···················································································································1分点P运动速度每秒钟1个单位长度．······················································································2分（2）过点B作BF⊥y轴于点F，BE⊥x轴于点E，则BF＝8，4OFBE．∴1046AF．在Rt△AFB中，228610AB3分过点C作CG⊥x轴于点G，与FB的延长线交于点H．∵90,ABCABBC∴△ABF≌△BCH．∴6,8BHAFCHBF．∴8614,8412OGFHCG．∴所求C点的坐标为（14，12）．4分（3）过点P作PM⊥y轴于点M，PN⊥x轴于点N，则△APM∽△ABF．∴APAMMPABAFBF．1068tAMMP．∴3455AMtPMt，．∴3410,55PNOMtONPMt．设△OPQ的面积为S（平方单位）∴213473(10)(1)5251010Stttt（0≤t≤10）·································································5分说明:未注明自变量的取值范围不扣分．∵310a0∴当474710362()10t时，△OPQ的面积最大．···································6分此时P的坐标为（9415，5310）．····························································································7分（4）当53t或29513t时，OP与PQ相等．··································································9分对一个加1分，不需写求解过程．例2．（1）A（8，0）B（0，6）················1分（2）86OAOB，10AB点Q由O到A的时间是881（秒）点P的速度是61028（单位/秒）··1分当P在线段OB上运动（或03t≤≤）时，2OQtOPt，2St············································································································································1分ABCDEFGHMNPQOxyxAOQPByD当P在线段BA上运动（或38t≤）时，6102162OQtAPtt，,如图，作PDOA于点D，由PDAPBOAB，得4865tPD，········································1分21324255SOQPDtt·····························································································1分（自变量取值范围写对给1分，否则不给分．）（3）82455P，··························································································································1分12382412241224555555IMM，，，，，·····································································3分注：本卷中各题，若有其它正确的解法，可酌情给分．例3（1）过点A作AEx⊥轴，垂足为E（如图1）(34)A，，22435AEOEOAAEOE，，．四边形ABCO为菱形，5OCCBBAOA，(50)C，．·············································································1分设直线AC的解析式为：ykxb，5034kbkb1252kb直线AC的解析式为：1522yx．···············································································1分（2）由（1）得M点的坐标为502，，52OM．如图1，当点P在AB边上运动时，由题意得342OHHM，．113(52)222SBPMHt31550242Stt≤．····························2分当P点在BC边上运动时，记为1P．OCMBCMCOCBCMCM，，，OMBHACxy图1EP1POMCBMC△≌△，52OMBM，90MOCMBC°．1115(25)222SPBBMt，52555242Stt≤．·······································2分（3）设OP与AC相交于点Q，连接OB交AC于点K．AOCABC，AOMABM．90MPBBCO°，BAOBCO，90BAOAOH°，MPBAOH，MPBMBH．当P点在AB边上运动时，如图2．MPBMBH，PMBM．MHPB⊥，2PHHB，1PAAHPH，12t．·························································1分ABOC∥，PAQOCQ．AQPCQO，AQPCQO△∽△，15AQAPCQCO．在RtAEC△中，22224845ACAEEC，253AQ，1053QC．在RtOHB△中，22222425OBHBHO，ACOBOKKBAKCK，，，525OKAKKC，，453QKAKAQ，3tan4OKOQCQK．·······································································································1分当P点在BC边上运动时，如图3，90BHMPBM°，MPBMBH，tantanMPBMBH，53222BMHMBPHBBP，，103BP，256t．·································1分OMBHACxy图2EKPQOMBHACxy图3KPQ105533PCBCBP．由PCOA∥，同理可证PQCOQA△∽△，CQCPAQAO．115534CQCQACQKKCCQAQ，，．5OK，tan1OKOQKKQ．···············································································1分综上所述，当12t时，MPB与BCO互为余角，直线OP与直线AC所夹锐角的正切值为34；当256t时，MPB与BCO互为余角，直线OP与直线AC所夹锐角的正切值为1．（以上各题如有不同解法并且正确，请按相应步骤给分）例4．解:(1)由题知:033903baba……………………………………1分解得:21ba……………………………………………………………2分∴所求抛物线解析式为:322x－－xy……………………………3分(2)存在符合条件的点P,其坐标为P(－1,10)或P(－1,－10)或P(－1,6)或P(－1,35)………………………………………………………7分(3)解法①：过点E作EF⊥x轴于点F,设E(a,-2a-2a＋3)(－3a0)∴EF=-2a-2a＋3，BF=a＋3，OF=－a………………………………………………8分∴S四边形BOCE=21BF·EF+21(OC+EF)·OF=21(a＋3)·(－2a－2a＋3)+21(－2a－2a＋6)·（－a）……………………………9分=2929232aa………………………………………………………………………10分=－232)23(a+863∴当a=－23时，S四边形BOCE最大,且最大值为863．……………………………11分此时，点E坐标为(－23，415)……………………………………………………12分解法②：过点E作EF⊥x轴于点