2009年全国各地中考数学压轴题参考答案及评分标准(三)

2009年全国各地中考数学压轴题参考答案及评分标准（三）201．解：（1）由题意得6＝a(－2＋3)(－2－1)，∴a＝－2．···································1分∴抛物线的解析式为y＝－2(x＋3)(x－1)，即y＝－2x2－4x＋6令－2(x＋3)(x－1)＝0，得x1＝－3，x2＝1∵点A在点B右侧，∴A(1，0)，B(－3，0)设直线AC的函数关系式为y＝kx＋b，把A(1，0)、C(－2，6)代入，得620＝＋＝＋－bkbk解得22＝＝－bk∴直线AC的函数关系式为y＝－2x＋2．···················3分（2）①设P点的横坐标为m(－2≤m≤1)，则P(m，－2m＋2)，M(m，－2m2－4m＋6)．················4分∴PM＝－2m2－4m＋6－(－2m＋2)＝－2m2－2m＋4＝－2(m＋21)2＋29∴当m＝－21时，线段PM长度的最大值为29．··········6分②存在M1(0，6)．···········································································7分M2(－41，855)．····································································9分点M的坐标的求解过程如下（原题不作要求，本人添加，仅供参考）ⅰ)如图1，当M为直角顶点时，连结CM，则CM⊥PM，△CMP∽△ANP∵点C(－2，6)，∴点M的纵坐标为6，代入y＝－2x2－4x＋6得－2x2－4x＋6＝6，∴x＝－2（舍去）或x＝0∴M1(0，6)（此时点M在y轴上，即抛物线与y轴的交点，此时直线MN与y轴重合，点N与原点O重合）ⅱ)如图2，当C为直角顶点时，设M(m，－2m2－4m＋6)(－2≤m≤1)过C作CH⊥MN于H，连结CM，设直线AC与y轴相交于点D则△CMP∽△NAP又∵△HMC∽△CMP，△NAP∽△OAD，∴△HMC∽△OAD∴ODCH＝OAMH∵C(－2，6)，∴CH＝m＋2，MH＝－2m2－4m＋6－6＝－2m2－4mOABxyCP图1MOABxyCPMN在y＝－2x＋2中，令x＝0，得y＝2∴D(0，2)，∴OD＝2∴22m＝1422mm整理得4m2＋9m＋2＝0，解得m＝－2（舍去）或m＝－41当m＝－41时，－2m2－4m＋6＝(－41)2－4×(－41)＋6＝855∴M2(－41，855)202．解：（1）∵抛物线抛物线y＝x2＋bx＋c经过A（1，0），B（0，2）∴ccb＋＋＝＋＋＝00210解得23＝＝－cb∴所求抛物线的解析式为y＝x2－3x＋2．·········································2分（2）∵A（1，0），B（0，2），∴OA＝1，OB＝2可得旋转后C点的坐标为（3，1）．···············································3分当x＝3时，由y＝x2－3x＋2，得y＝2可知抛物线y＝x2－3x＋2过点（3，2）．∴将原抛物线沿y轴向下平移1个单位后过点C∴平移后所得图象的函数关系式为y＝x2－3x＋1．·····························5分（3）∵点N在y＝x2－3x＋1上，∴可设点N的坐标为（x0，x02－3x0＋1）∵y＝x2－3x＋1＝(x－23)2－45，∴其对称轴为x＝23．······················6分①当0＜x0＜23时，如图①∵S△NBB1＝2S△NDD1∴21×1×x0＝2×21×1×(23－x0)∴x0＝1此时x02－3x0＋1＝－1∴点N的坐标为（1，－1）．······························8分当x0＞23时，如图②同理可得21×1×x0＝2×21×1×(x0－23)∴x0＝3此时x02－3x0＋1＝1∴点N的坐标为（3，1）OABxyCPN图2MDHyBOADx图①D1CB1NyBOADx图②D1CB1N综上，点N的坐标为（1，－1）或（3，1）．·······10分203．解：（1）把x＝0代入y＝－x－3，得y＝－3，∴C（0，－3）∵抛物线y＝ax2＋bx＋c过点C（0，－3），∴c＝－3∵抛物线的的顶点D（ab2－，abac442－）在直线y＝－x－3上∴aba4342－－)(＝ab2－3，解得b＝0或b＝－2···········································2分当b＝0时，点C、D重合，不合题意，舍去∴b＝－2，∴D（a1，－3－a1）如图1，过D作DH⊥y轴于H，在Rt△CHD中，CD2＝CH2＋HD2即(2)2＝[－3－(－3－a1)]2＋(a1)2，解得a＝1或a＝－1当a＝－1时，△＝b2－4ac＝(－2)2－4×(－1)×(－3)＝－8＜0，函数图象与x没有交点，不合题意，舍去．·········································································3分∴所求抛物线的解析式为y＝x2－2x－3．·········································4分（2）设直线MN的解析式为y＝k，代入y＝x2－2x－3，得x2－2x－3＝k即x2－2x－3－k＝0，解得x1＝1＋4＋k，x2＝1－4＋k（k＞－3）∴线段MN的长为42＋k∵以MN为直径的圆与x轴相切，∴4＋k＝|k|．···························5分整理得k2－k－4＝0，解得k＝2171＋或k＝2171－．·······················6分设该圆的半径为r，则当k＝2171＋时，直线MN在x轴的上方∴r＝k＝2171＋．········································7分当k＝2171－时，直线MN在x轴的下方∴r＝－k＝2171＋－．··································8分（3）在y＝x2－2x－3中，令y＝0，得x1＝－1，x2＝3∵A在B的左侧，∴A（－1，0），B（3，0）∴OA＝1，OB＝3在y＝－x－3中，令y＝0，得x＝－3∴E（－3，0），∴OE＝OC＝3∵AF∥EC，∴△AOF∽△EOCOABxyCD图1HOABxyCDM图2NMN∴OFOA＝OCOE＝1，∴OF＝OA＝1，∴F（0，－1）∴S四边形AECF＝S△EOC－S△AOF＝21×3×3－21×1×1＝4．······················9分又A（－1，0），可得直线AE的解析式为y＝－x－1如图3，设直线PB分别与AF、EC交于点G、H，设G（m，－m－1）易得直线PB的解析式为y＝mm－＋31x－mm－＋333（－1＜m＜0）联立333331－－－－－＝＋＋＝xmmxmmyy解得233233＋＝＝－－mmxy∴H（233－m，233＋－m）S四边形AEHG＝S△BEH－S△BAG＝21×6×233＋m－21×4×(m＋1)＝255＋m若直线PB将四边形AECF的面积平分，则S四边形AEHG＝21S四边形AECF＝2即255＋m＝2，解得m＝－51∵OE＝OC，OA＝OF，∴AE＝FC∴四边形AECF是等腰梯形，设其高为h当S四边形AEHG＝S四边形CFGH时，有21(AG＋EH)·h＝21(GF＋HC)·h即AG＋EH＝GF＋HC∴AG＋EH＋AE＝GF＋HC＋FC∴此时直线PB也将四边形AECF的周长平分．·······························10分当m＝－51时，直线PB的解析式为y＝41x－43联立3243412－－－＝＝xxxyy解得16154311－－＝＝yx0322＝＝yx∵B（3，0），∴P（－43，－1615）··················································11分综上所述，在抛物线上存在点P（－43，－1615），使直线PB恰好将四边形AECF的周长和面积同时平分．············································································12分OABxyCDF图3EGHP204．解：（1）如图1，过B作BD⊥OA于D∵B（7，4），tan∠BAO＝34∴OD＝7，BD＝4，AD＝3．·················1分∴OA＝OD＋AD＝10∴点A的坐标为（10，0）．··················2分∵等腰梯形OABC∴点C的坐标为（3，4）·····················3分（2）∵抛物线经过O、B、C三点∴设抛物线的解析式为y＝ax2＋bx（a≠0），把B（7，4），C的坐标为（3，4）代入，得baba3947494＋＝＋＝．·········································································4分解得2140214＝＝－ba．········································································5分故所求抛物线的解析式为y＝－214x2＋2140x．···································6分（3）设经过点P且与等腰梯形一腰平行的直线分别交BC、OA于点E、FBC＝7－3＝4①如图2，当EF∥OC时，则四边形OFEC是平行四边形，设OF＝x若S□OFEC＝21S梯形OABC则有4x＝21×21(4＋10)×4解得x＝27，∴F（27，0）．······················7分∴E点的横坐标＝27＋3＝213，∴E（213，4）········································8分设直线EF的解析式为y＝kx＋b（k≠0）则bkbk＋＝＋＝2702134解得31434－＝＝bk∴直线EF的解析式为y＝34x－314．··············································9分由34x－314＝－214x2＋2140x，得x＝21073∴在第一象限内，所求点P的横坐标为21073＋．···························10分②如图3，当EF∥AB时，则四边形ABEF是平行四边形ABxyCO图2PFEABxyCO图1D若S□ABEF＝21S梯形OABC由①知AF＝27，∴OF＝10－27＝213，∴F（213，0）∴E点的横坐标＝213－3＝27，∴E（27，4）设直线EF的解析式为y＝mx＋n（m≠0）则nmnm＋＝＋＝2130274解得32634＝＝－nm∴直线EF的解析式为y＝－34x＋326．·········································11分由－34x＋326＝－214x2＋2140x，得x＝210717∴在第一象限内，所求点P的横坐标为210717－．·························12分205．证明：（1）∵BC是⊙O的直径，∴∠BAC＝90°又∵ME⊥BC，BM平分∠ABC，∴AM＝ME，∠AMN＝∠EMN又∵MN＝MN，∴△△ANM≌△ENM．········································3分（2）∵AB2＝AF·AC，∴ACAB＝ABAF又∵∠BAC＝∠FA