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)

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

6，那么5EF=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．

y

26．解：（1）由已知，得C(3，0)，D(2，2)，

A E x

D B O 26题图

C ??ADE?90°??CDB??BCD，

1?AE?AD?tan?ADE?2?tan?BCD?2??1．

2···················································································································· （1分） ，． ·?E(01)设过点E、D、C的抛物线的解析式为y?ax?bx?c(a?0)． 将点E的坐标代入，得c?1．

将c?1和点D、C的坐标分别代入，得

2?4a?2b?1?2， ············································································································ （2分） ??9a?3b?1?0.5?a????6解这个方程组，得?

13?b??6?故抛物线的解析式为y??5213······························································· （3分） x?x?1． ·

661

（2）EF?2GO成立． ································································································ （4分）

6?点M在该抛物线上，且它的横坐标为，

512······························································································ （5分） ?点M的纵坐标为． ·5设DM的解析式为y?kx?b1(k?0)， 将点D、M的坐标分别代入，得

F A E x

y Ｍ D B 1?2k?b1?2，???k??，2 ?612 解得?k?b1?.??5?b1?3．?51··············································································· （6分） ?DM的解析式为y??x?3．

2·································································································· （7分） 3)，EF?2． ·?F(0，过点D作DK⊥OC于点K，

则DA?DK．

??ADK??FDG?90°， ??FDA??GDK．

又??FAD??GKD?90°， ?△DAF≌△DKG． ?KG?AF?1．

··················································································································· （8分） ?GO?1． ·

?EF?2GO．

（3）?点P在AB上，G(1，0)，C(3，0)，则设P(1，2)．

O G K C ?PG2?(t?1)2?22，PC2?(3?t)2?22，GC?2．

①若PG?PC，则(t?1)?2?(3?t)?2， 解得t?2．?P(2，2)，此时点Q与点P重合．

··················································································································· （9分） 2)． ·?Q(2，②若PG?GC，则(t?1)?2?2， 解得 t?1，?P(1，2)，此时GP⊥x轴．

2?22222GP与该抛物线在第一象限内的交点Q的横坐标为1，

7?点Q的纵坐标为．

3 2

?7?··············································································································· （10分） ?Q?1，?． ·

?3?③若PC?GC，则(3?t)?2?2，

解得t?3，?P(3，2)，此时PC?GC?2，△PCG是等腰直角三角形． 过点Q作QH⊥x轴于点H， 则QH?GH，设QH?h，

y Q A E P Q (Q) (P) D B (P) 222?Q(h?1，h)．

513??(h?1)2?(h?1)?1?h．

667解得h1?，h2??2（舍去）．

5?127?·············································· （12分） ?Q?，?． ·

55??综上所述，存在三个满足条件的点Q，

O G H C x

即Q(2，2)或Q?1，?或Q?

?7??3??127?，?． ?55?3、（2009年重庆綦江县）26．（11分）如图，已知抛物线y?a(x?1)2?33(a?0)经过点A(?2，0)，抛物线的顶点为D，过O作射线OM∥AD．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t(s)．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OC?OB，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？

M y D 并求出最小值及此时PQ的长． C P A O Q B x 3

*26．解：（1）?抛物线y?a(x?1)?33(a?0)经过点A(?2，0)，

2?0?9a?33?a??3 ·································································································· 1分 3322383x?x? ······················································· 3分 333?二次函数的解析式为：y??33)过D作DN?OB于N，则DN?33， （2）?D为抛物线的顶点?D(1，AN?3，?AD?32?(33)2?6??DAO?60° ························································ 4分

?OM∥AD

①当AD?OP时，四边形DAOP是平行四边形

···················································· 5分 ?OP?6?t?6(s) ·

y D M C ②当DP?OM时，四边形DAOP是直角梯形

A H P B x 过O作OH?AD于H，AO?2，则AH?1 O E N Q （如果没求出?DAO?60°可由Rt△OHA∽Rt△DNA求AH?1） ···································································································· 6分 ?OP?DH?5t?5(s) ·

③当PD?OA时，四边形DAOP是等腰梯形

?OP?AD?2AH?6?2?4?t?4(s)

综上所述：当t?6、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，?COB?60°，OC?OB，△OCB是等边三角形 则OB?OC?AD?6，OP?t，BQ?2t，?OQ?6?2t(0?t?3)

过P作PE?OQ于E，则PE?3t ················································································ 8分 2113?SBCPQ??6?33??(6?2t)?t

2223?3?633 =··········································································································· 9分 ?t???2?2?8当t?2363时，SBCPQ的面积最小值为········································································· 10分 3 28 4