冲刺2010—2009年中考数学压轴题汇编(含解题过程)

冲刺2010 ——2009年中考数学压轴题汇编(含解题过程)

1、（2009年北京）25.如图，在平面直角坐标系xOy中，ABC三个机战的坐标分别为

6,0A，6,0B，0,43C，延长AC到点D,使CD=12AC,过点D作DE∥AB交BC的延长线于点E.

（1）求D点的坐标；

（2）作C点关于直线DE的对称点F,分别连结DF、EF，若过B点的直线ykxb将四边形CDFE分成周长相等的两个四边形，确定此直线的解析式；

（3）设G为y轴上一点，点P从直线ykxb与y轴的交点出发，先沿y轴到达G点，再沿GA到达A点，若P点在y轴上运动的速度是它在直线GA上运动速度的2倍，试确定G点的位置，使P点按照上述要求到达A点所用的时间最短。（要求：简述确定G点位置的方法，但不要求证明）

2、（2009年重庆市）26．已知：如图，在平面直角坐标系xOy中，矩形OABC的边OA在y轴的正半轴上，OC在x轴的正半轴上，OA=2，OC=3．过原点O作∠AOC的平分线交AB于点D，连接DC，过点D作DE⊥DC，交OA于点E．

（1）求过点E、D、C的抛物线的解析式；

（2）将∠EDC绕点D按顺时针方向旋转后，角的一边与y轴的正半轴交于点F，另一边与线段OC交于点G．如果DF与（1）中的抛物线交于另一点M，点M的横坐标为65，那么EF=2GO是否成立？若成立，请给予证明；若不成立，请说明理由；

（3）对于（2）中的点G，在位于第一象限内的该抛物线上是否存在点Q，使得直线GQ与AB的交点P与点C、G构成的△PCG是等腰三角形？若存在，请求出点Q的坐标；若不存在，请说明理由．

26．解：（1）由已知，得(30)C，，(22)D，，

90ADECDBBCD°，

1tan2tan212AEADADEBCD．

(01)E，． ····················································································································· （1分）

设过点EDC、、的抛物线的解析式为2(0)yaxbxca．

将点E的坐标代入，得1c．

将1c和点DC、的坐标分别代入，得

42129310.abab， ············································································································ （2分）

解这个方程组，得56136ab

故抛物线的解析式为2513166yxx． ································································ （3分）

（2）2EFGO成立． ································································································ （4分） 26题图 y

x D B

C A

E

O

点M在该抛物线上，且它的横坐标为65，

点M的纵坐标为125． ······························································································· （5分）

设DM的解析式为1(0)ykxbk，

将点DM、的坐标分别代入，得

1122612.55kbkb， 解得1123kb，．

DM的解析式为132yx． ··············································································· （6分）

(03)F，，2EF． ··································································································· （7分）

过点D作DKOC⊥于点K，

则DADK．

90ADKFDG°，

FDAGDK．

又90FADGKD°，

DAFDKG△≌△．

1KGAF．

1GO． ···················································································································· （8分）

2EFGO．

（3）点P在AB上，(10)G，，(30)C，，则设(12)P，．

222(1)2PGt，222(3)2PCt，2GC．

①若PGPC，则2222(1)2(3)2tt，

解得2t．(22)P，，此时点Q与点P重合．

(22)Q，． ···················································································································· （9分）

②若PGGC，则22(1)22t，

解得 1t，(12)P，，此时GPx⊥轴．

GP与该抛物线在第一象限内的交点Q的横坐标为1，

点Q的纵坐标为73．

713Q，． ················································································································ （10分） y

x D B

C A

E

O Ｍ F

K G

③若PCGC，则222(3)22t，

解得3t，(32)P，，此时2PCGC，PCG△是等腰直角三角形．

过点Q作QHx⊥轴于点H，

则QHGH，设QHh，

(1)Qhh，．

2513(1)(1)166hhh．

解得12725hh，（舍去）．

12755Q，． ··············································· （12分）

综上所述，存在三个满足条件的点Q，

即(22)Q，或713Q，或12755Q，．

3、（2009年重庆綦江县）26．（11分）如图，已知抛物线(1)233(0)yaxa经过点(2)A，0，抛物线的顶点为D，过O作射线OMAD∥．过顶点D平行于x轴的直线交射线OM于点C，B在x轴正半轴上，连结BC．

（1）求该抛物线的解析式；

（2）若动点P从点O出发，以每秒1个长度单位的速度沿射线OM运动，设点P运动的时间为()ts．问当t为何值时，四边形DAOP分别为平行四边形？直角梯形？等腰梯形？

（3）若OCOB，动点P和动点Q分别从点O和点B同时出发，分别以每秒1个长度单位和2个长度单位的速度沿OC和BO运动，当其中一个点停止运动时另一个点也随之停止运动．设它们的运动的时间为t()s，连接PQ，当t为何值时，四边形BCPQ的面积最小？并求出最小值及此时PQ的长．

*26．解：（1）抛物线2(1)33(0)yaxa经过点(20)A，， y

x D B

C A

E

O Q

P

H G (P) (Q)

Q (P)

x y M

C D

P

Q O A

B

309333aa ·································································································· 1分

二次函数的解析式为：232383333yxx ······················································· 3分

（2）D为抛物线的顶点(133)D，过D作DNOB于N，则33DN，

2233(33)660ANADDAO，° ························································ 4分

OMAD∥

①当ADOP时，四边形DAOP是平行四边形

66(s)OPt ····················································· 5分

②当DPOM时，四边形DAOP是直角梯形

过O作OHAD于H，2AO，则1AH

（如果没求出60DAO°可由RtRtOHADNA△∽△求1AH）

55(s)OPDHt ····································································································· 6分

③当PDOA时，四边形DAOP是等腰梯形

26244(s)OPADAHt

综上所述：当6t、5、4时，对应四边形分别是平行四边形、直角梯形、等腰梯形． ·· 7分

（3）由（2）及已知，60COBOCOBOCB°，，△是等边三角形

则6262(03)OBOCADOPtBQtOQtt，，，

过P作PEOQ于E，则32PEt ················································································ 8分

113633(62)222BCPQStt

=233633228t ··········································································································· 9分

当32t时，BCPQS的面积最小值为6338 ········································································· 10分

此时3339333324444OQOPOEQEPE，=， x y M

C D

P

Q O A

B N E H