内容发布更新时间 : 2024/7/3 20:36:48星期一 下面是文章的全部内容请认真阅读。

2009年全国中考数学压轴题精选精析（重庆）

1.（2009年重庆綦江）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0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0出发，以每秒l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l个长度单位和2个长度单位的速度沿OC和B0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．

（2009年重庆綦江26题

2y?a(x?1)?3解析）解：（1）抛物线

A(?2，0)， 经过点3a(?0)?0?9a?33?a??3 ·································································································· 1分 3322383 ······················································· 3分 x?x?333?二次函数的解析式为：y??（2）

D为抛物线的顶点?D(13，3)过D作DN?OB于N，则DN?33，

AN?3，?AD?32?(33)2?6??DAO?60° ························································ 4分

OM∥AD

y ①当AD?OP时，四边形DAOP是平行四边形

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

D M C ②当DP?OM时，四边形DAOP是直角梯形

过O作OH?AD于H，AO?2，则AH?1

A H P B x O E N Q （如果没求出?DAO?60°可由Rt△OHA∽Rt△DNA求AH?1）

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

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

，OC?OB，△OCB是等边三角形 （3）由（2）及已知，?COB?60°?OQ?6?2t(0?t?3) 则OB?OC?AD?6，OP?t，BQ?2t，过P作PE?OQ于E，则PE?3··············································································· 8分 t ·2113?SBCPQ??6?33??(6?2t)?t

2223?3?63=·········································································································· 9分 3·?t???2?2?8当t?23633 时，SBCPQ的面积最小值为········································································· 10分

2833?此时OQ?3，OP=，OE?24

?QE?3?39?44PE?33 4

?33??9?23322 ··························································· 11分 ?PQ?PE?QE?????4????2???4?

2.（2009年重庆）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

O C x

26题图

0)，D(2，2)， （2009年重庆26题解析）解：（1）由已知，得C(3，A E D B 26，那么EF=2GO是否成立？若成立，请5?ADE?90°??CDB??BCD，

?AE?ADtan?ADE?2?tan?BCD?2?1?1． 21)． ····················································································································· （1分） ?E(0，

设过点E、D、C的抛物线的解析式为y?ax2?bx?c(a?0)． 将点E的坐标代入，得c?1．

[来源学&科&网]

将c?1和点D、C的坐标分别代入，得

?4a?2b?1?2， ············································································································ （2分） ?9a?3b?1?0.?5?a????6解这个方程组，得??b?13?6?故抛物线的解析式为y??

[来源学#科#网]5213x?x?1． ································································ （3分） 666， 5y （2）EF?2GO成立． ································································································ （4分） 点M在该抛物线上，且它的横坐标为

?点M的纵坐标为

12． ······························································································· （5分） 5F A E x

设DM的解析式为y?kx?b1(k?0)， 将点D、M的坐标分别代入，得

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

23)，EF?2． ··································································································· （7分） ?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．[来源学。科。网Z。X。X。K]

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

?EF?2GO．

（3）

点P在AB上，G(1，0)，C(3，0)，则设P(1，2)．

?PG2?(t?1)2?22，PC2?(3?t)2?22，GC?2．

①若PG?PC，则(t?1)2?22?(3?t)2?22，

解得t?2．?P(2，2)，此时点Q与点P重合． ?Q(2，2)． ···················································································································· （9分） ②若PG?GC，则(t?1)2?2??22， 解得 t?1，?P(1，2)，此时GP⊥x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．

?Q??7??1，3??． ·

··············································································································· （10分） ③若PC?GC，则(3?t)2?22?22，

[来源:Zxxk.Com]

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，

y Q (QD )( P)

A P B (P) E Q

则QH?GH，设QH?h，

?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，

2)或Q?1，?或Q?即Q(2，

?7??3??127?，?． ?55?