5、2011中考压轴之四边形存在性问题 - 图文

1、(2009年黑龙江齐齐哈尔) 直线y??3x?6与坐标轴分别交于A、B两点，动点P、Q4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时，求出点P的坐标，并直接写出以点O、P、Q为顶点的平行四边形的第5y B 四个顶点M的坐标．

（1）A（8，0）B（0，6） ························ 1分 （2）?OA?8，OB?6 ?AB?10

8?点Q由O到A的时间是?8（秒）

16?10?2（单位/秒） ··· 1分 ?点P的速度是8P x O Q A 当P在线段OB上运动（或0≤t≤3）时，OQ?t，OP?2t

S?t2 ······································································································································ 1分

当P在线段BA上运动（或3?t≤8）时，OQ?t，AP?6?10?2t?16?2t, 如图，作PD?OA于点D，由

PDAP48?6t?，得PD?， ······································ 1分 BOAB51324?S?OQ?PD??t2?t ························································································· 1分

255（自变量取值范围写对给1分，否则不给分．）

（3）P?，? ····················································································································· 1分

?824??55???824??1224??1224?································································· 3分 I1?，?，M2??，?，M3?，?? ·

555555??????2、（2010河南）在平面直角坐标系中，已知抛物线经过A(?4,0)，B(0,?4)，C(2,0)三点．

（1）求抛物线的解析式；

（2）若点M为第三象限内抛物线上一动点，点M的横坐标为m，△AMB的面积为S．求S关于m的函数关系式，并求出S的最大值．

（3）若点P是抛物线上的动点，点Q是直线y??x上的动点，判断有几个位置能够使得点P、Q、B、O为顶点的四边形为平行四边形，直接写出相应的点Q的坐标．

yAOCxMB

3、（2011黑龙江鸡西）已知直线y=3x＋43与x轴，y轴分别交于A、B两点， ∠ABC=60°，BC与x轴交于点C. （1）试确定直线BC的解析式.

（2）若动点P从A点出发沿AC向点C运动（不与A、C重合），同时动点Q从C点出发

沿CBA向点A运动(不与C、A重合) ，动点P的运动速度是每秒1个单位长度，动点Q的运动速度是每秒2个单位长度.设△APQ的面积为S，P点的运动时间为t秒，求S与t的函数关系式，并写出自变量的取值范围.

（3）在（2）的条件下，当△APQ的面积最大时，y轴上有一点M，平面内是否存在一点N，

使以A、Q、M、N为顶点的四边形为菱形？若存在，请直接写出A N点的坐标；若不存在，请说明理由.

解：( 1 )由已知得A点坐标(－4﹐0)，B点坐标(0﹐43﹚

∵OA＝4 OB＝43 ∴∠BAO＝60o ∵∠ABC＝60o ∴△ABC是等边三角形 ∵OC＝OA＝4 ∴C点坐标﹙4，0﹚

设直线BC解析式为y＝kx﹢b

??b?43?4k?b?0 ∴?k??3 ????b?43∴直线BC的解析式为y=－3x?43 ------------------------------------------ (2分) ﹙2﹚当P点在AO之间运动时，作QH⊥x轴。

∵QHCQOB?CB ∴QH43?2t8 ∴QH=3t ∴S△APQ=

12AP·QH=12t·3t=32t2（0＜t≤4）---------------------------------------（2分） 同理可得S1△APQ=

2t·﹙83?3t﹚=－32t2?43t﹙4≤t＜8﹚--------------（2分） （3）存在，（4，0），（－4，8）（－4，－8）（－4，833） ----------------------（4分）

Q

P

H

Q 4、（2007河南）如图，对称轴为直线x＝

7的抛物线经过点A（6，0）和B（0，4）． 2（1）求抛物线解析式及顶点坐标；

（2）设点E（x，y）是抛物线上一动点，且位于第四象限，四边形OEAF是以OA为对角线的平行四边形，求四边形OEAF的面积S与x之间的函数关系式，并写出自变量x的取值范围；

（3）①当四边形OEAF的面积为24时，请判断OEAF是否为菱形？

②是否存在点E，使四边形OEAF为正方形？若存在，求出点E的坐标；若不存

在，请说明理由．

yx=72B(0,4)FOEA(6,0)x

5、(2010年大兴安岭) 如图，在平面直角坐标系中，函数y＝2x＋12的图象分别交x轴、y

轴于A、B两点．过点A的直线交y轴正半轴于点M，且点M为线段OB的中点．△ABP△AOB

（1）求直线AM的解析式；

（2）试在直线AM上找一点P，使得S△ABP＝S△AOB ，请直接写出点P的坐标；

（3）若点H为坐标平面内任意一点，在坐标平面内是否存在这样的点H，使以A、B、

M、H为顶点的四边形是等腰梯形？若存在，请直接写出点H的坐标；若不存在，请说明理由．

解：（1）函数的解析式为y＝2x＋12 ∴A(－6,0)，B(0,12) ………………1分

∵点M为线段OB的中点 ∴M(0,6) ……………………………1分 设直线AM的解析式为：y＝kx＋b

b＝6 ∵ ………………………………………………2分

－6k＋b＝0 ∴k＝1 b＝6 ………………………………………………………1分 ∴直线AM的解析式为：y＝x＋6 ………………………………………1分 （2）P1(－18，－12)，P2(6，12) ………………………………………………2分

618

（3）H1(－6，18)，H2(－12，0)，H3(－ ， )………………………………3分

556、（2009抚顺）已知：如图所示，关于x的抛物线y?ax?x?c(a?0)与x轴交于点

2A(?2，0)、点B(6，0)，与y轴交于点C．

（1）求出此抛物线的解析式，并写出顶点坐标；

（2）在抛物线上有一点D，使四边形ABDC为等腰梯形，写出点D的坐标，并求出直线AD的解析式；

（3）在（2）中的直线AD交抛物线的对称轴于点M，抛物线上有一动点P，x轴上有一动点Q．是否存在以A、M、P、Q为顶点的平行四边形？如果存在，请直接写出点Q的坐标；如果不存在，请说明理由． 解：（1）根据题意，得

C y A O B x （第26题图）

C y D P1 Q1 O Q3 第26题图

Q4 B P4 x ?4a?2?c?0 ·············································· 1分 ??36a?6?c?0C P2 A 1??a??解得?·················································· 3分 4 ·

??c?31········· 4分 ?抛物线的解析式为y??x2?x?3 ·

4Q2 P3 顶点坐标是（2，4） ··············································································································· 5分

3) ·（2）D(4，························································································································· 6分

设直线AD的解析式为y?kx?b(k?0)

0)点D(4，3) ?直线经过点A(?2，、??2k?b?0 ······················································································································ 7分 ??4k?b?3?1?k????······························································································································· 8分 2 ·??b?1?y?1x?1 ·························································································································· 9分 2（3）存在． ·························································································································· 10分 ······················································································································ 11分 Q1(22?2，0) ··············································································································· 12分 Q2(?22?2，0) ·

······················································································································ 13分 Q3(6?26，0) ······················································································································ 14分 Q4(6?26，0)

（3）在（2）中的直线AD交抛物线的对称轴于点M，抛物线上有一动点P，x轴上有一动点Q．是否存在以A、M、P、Q为顶点的平行四边形？如果存在，请直接写出点Q的坐标；如果不存在，请说明理由． 解：（1）根据题意，得

C y A O B x （第26题图）

C y D P1 Q1 O Q3 第26题图

Q4 B P4 x ?4a?2?c?0 ·············································· 1分 ??36a?6?c?0C P2 A 1??a??解得?·················································· 3分 4 ·

??c?31········· 4分 ?抛物线的解析式为y??x2?x?3 ·

4Q2 P3 顶点坐标是（2，4） ··············································································································· 5分

3) ·（2）D(4，························································································································· 6分

设直线AD的解析式为y?kx?b(k?0)

0)点D(4，3) ?直线经过点A(?2，、??2k?b?0 ······················································································································ 7分 ??4k?b?3?1?k????······························································································································· 8分 2 ·??b?1?y?1x?1 ·························································································································· 9分 2（3）存在． ·························································································································· 10分 ······················································································································ 11分 Q1(22?2，0) ··············································································································· 12分 Q2(?22?2，0) ·

······················································································································ 13分 Q3(6?26，0) ······················································································································ 14分 Q4(6?26，0)

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