2013年中考数学压轴题精选(二次函数)(16题)_附详细解答和评分标准

如图，在平面直角坐标系中，抛物线y=－

22x+bx+c经3y 过A（0，－4）、B（x1，0）、 C（x2，0）三点，且x2-x1=5． （1）求b、c的值；（4分）

（2）在抛物线上求一点D，使得四边形BDCE是以BC为对 角线的菱形；（3分）

（3）在抛物线上是否存在一点P，使得四边形BPOH是以OB为对角线的菱形？若存在，求出点P的坐标，并判断这个菱形是否为正方形？若不存在，请说明理由．（3分）

解：

（08广东茂名25题解析）解：（1）解法一： ∵抛物线y=－

B C O x

A （第25题图）

22， x+bx+c经过点A（0，－4）

322x+bx+c=0的两个根， 3 ∴c=－4 „„1分

又由题意可知，x1、x2是方程－∴x1+x2=

33b， x1x2=－c=6 ·········································································· 2分 222由已知得（x2-x1）=25 又（x2-x∴

1）=（x2+x1）－4x122x2=

92b－24 492b－24=25 414解得b=± ···················································································································· 3分

314当b=时，抛物线与x轴的交点在x轴的正半轴上，不合题意，舍去．

3∴b=－

14． ··················································································································· 4分 3解法二：∵x1、x2是方程－

即方程2x－3b∴x=

222x+bx+c=0的两个根， 3x+12=0的两个根．

3b9b296， ·················································································· 2分

4 1

9b296∴x2－x1==5，

2 解得 b=±14 ········································································································ 3分 3 （以下与解法一相同．）

（2）∵四边形BDCE是以BC为对角线的菱形，根据菱形的性质，点D必在抛物线的

对称轴上， ··········································································································· 5分

22142725···································· 6分 x－x－4=－（x+）2+ ·

33326725 ∴抛物线的顶点（－，）即为所求的点D． ·········································· 7分

26 又∵y=－

（3）∵四边形BPOH是以OB为对角线的菱形，点B的坐标为（－6，0），

根据菱形的性质，点P必是直线x=-3与

2214································································· 8分 x-x-4的交点， ·

332142 ∴当x=－3时，y=－×（－3）－×（－3）－4=4，

33抛物线y=－

∴在抛物线上存在一点P（－3，4），使得四边形BPOH为菱形． ··················· 9分 四边形BPOH不能成为正方形，因为如果四边形BPOH为正方形，点P的坐标

只能是（－3，3），但这一点不在抛物线上． ························································· 10分 2、（08广东肇庆25题）（本小题满分10分）

已知点A（a，y1）、B（2a，y2）、C（3a，y3）都在抛物线y5x212x上. （1）求抛物线与x轴的交点坐标； （2）当a=1时，求△ABC的面积；

（3）是否存在含有y1、y2、y3，且与a无关的等式？如果存在，试给出一个，并加以证明；如果不存在，说明理由.

（08广东肇庆25题解析）（本小题满分10分）

解：（1）由5x12x=0， ····················································································· （1分）

212． ························································································· （2分） 512∴抛物线与x轴的交点坐标为（0，0）、（，0）． ·········································· （3分）

5得x10，x2（2）当a=1时，得A（1，17）、B（2，44）、C（3，81）， ································· （4分） 分别过点A、B、C作x轴的垂线，垂足分别为D、E、F，则有

······················································· （5分） SABC=S梯形ADFC -S梯形ADEB -S梯形BEFC ·

2

=

(1781)2(1744)1(4481)1-- ······································· （6分）

222=5（个单位面积） ·············································································· （7分）

（3）如：y33(y2y1)． ················································································ （8分）

事实上，y35(3a)212(3a) =45a2+36a． 3（y2y1）=3[5×（2a）2+12×2a-（5a2+12a）] =45a2+36a．·············· （9分） ∴y33(y2y1)． ··························································································· （10分） 3、（08辽宁沈阳26题）（本题14分）26．如图所示，在平面直角坐标系中，矩形ABOC的边BO在x轴的负半轴上，边OC在y轴的正半轴上，且AB1，OB3，矩形ABOC绕点O按顺时针方向旋转60后得到矩形EFOD．点A的对应点为点E，点B的对应点为点F，点C的对应点为点D，抛物线yax2bxc过点A，E，D．

（1）判断点E是否在y轴上，并说明理由； （2）求抛物线的函数表达式；

（3）在x轴的上方是否存在点P，点Q，使以点O，B，P，Q为顶点的平行四边形的面积是矩形ABOC面积的2倍，且点P在抛物线上，若存在，请求出点P，点Q的坐标；若不存在，请说明理由．

（08辽宁沈阳26题解析）解：（1）点E在y轴上························································ 1分 理由如下：

连接AO，如图所示，在Rt△ABO中，AB1，BO3，AO2

B O 第26题图 A F C D x

y E sinAOB1，AOB30 2由题意可知：AOE60

BOEAOBAOE306090

·············································································· 3分 点B在x轴上，点E在y轴上． ·（2）过点D作DMx轴于点M

OD1，DOM30

3

在Rt△DOM中，DM点D在第一象限，

13，OM 2231································································································· 5分 点D的坐标为2， ·2由（1）知EOAO2，点E在y轴的正半轴上

2) 点E的坐标为(0，··································································································· 6分 点A的坐标为(31)， ·

抛物线yax2bxc经过点E，

c2

312，由题意，将A(31)代入yaxbx2中得 ，，D2283a3b21a9 解得 331b2ab534229853x2····························································· 9分 所求抛物线表达式为：yx299（3）存在符合条件的点P，点Q． ················································································ 10分 理由如下：矩形ABOC的面积ABBO3 以O，B，P，Q为顶点的平行四边形面积为23．

由题意可知OB为此平行四边形一边， 又OB3

OB边上的高为2 ··········································································································· 11分

2) 依题意设点P的坐标为(m，853x2上 点P在抛物线yx299853m2m22

99

4

解得，m10，m253 8532P2)，P21(0，8，

以O，B，P，Q为顶点的四边形是平行四边形，

PQ∥OB，PQOB3， 2)时， 当点P1的坐标为(0，点Q的坐标分别为Q1(3，2)，Q2(3，2)； 当点P2的坐标为A B F C D x O M y E 5328，时，

13333，22点Q的坐标分别为Q3··················································· 14分

，Q48，． ·8

4、（08辽宁12市26题）（本题14分）26．如图16，在平面直角坐标系中，直线y3x3与x轴交于点A，与y轴交于点C，抛物线yax223xc(a0)3y 经过A，B，C三点．

（1）求过A，B，C三点抛物线的解析式并求出顶点F的坐标；

（2）在抛物线上是否存在点P，使△ABP为直角三角形，若存在，直接写出P点坐标；若不存在，请说明理由； （3）试探究在直线AC上是否存在一点M，使得△MBF的周长最小，若存在，求出M点的坐标；若不存在，请说明理由．

（08辽宁12市26题解析）

解：（1）直线y3x3与x轴交于点A，与y轴交于点C．

A C O F B x

图16 A(1，0)，C(0，································································································ 1分 3) ·

点A，C都在抛物线上，

2330aca 3 3c33c抛物线的解析式为y

3223xx3 ······························································ 3分 335

431， ······································································································· 4分 顶点F3（2）存在 ························································································································· 5分 ······················································································································ 7分 P，3)·1(0····················································································································· 9分 P，3) ·2(2（3）存在 ······················································································································· 10分

理由： 解法一：

BC，延长BC到点B，使BC连接BF交直线AC于点M，则点M就是所求的点．

····························································································· 11分 过点B作BHAB于点H．

y B点在抛物线y32230) xx3上，B(3，33H A C B O B x

3在Rt△BOC中，tanOBC，

3OBC30，BC23，

在Rt△BBH中，BHM F 图9 1BB23， 2··················································· 12分 BH3BH6，OH3，B(3，23) ·设直线BF的解析式为ykxb

3233kbk643 解得

kbb3332y333x ········································································································· 13分 623 3y3x3x3107M， 解得 33377yxy103，6273103． ··· 14分 在直线AC上存在点M，使得△MBF的周长最小，此时M7，7

6

5、（08青海西宁28题）如图14，已知半径为1的O1与x轴交于A，B两点，OM为O10)，二次函数yx2bxc的图象经过A，B的切线，切点为M，圆心O1的坐标为(2，两点．

（1）求二次函数的解析式；

（2）求切线OM的函数解析式；

（3）线段OM上是否存在一点P，使得以P，O，A为顶点的三角形与△OO1M相似．若存在，请求出所有符合条件的点P的坐标；若不存在，请说明理由．

（08青海西宁28题解析）解：（1）0)，O1圆心O1的坐标为(2，图14

O A O1 B x y M ，0)，B(3，0)„„1分 半径为1，A(1二次函数yx2bxc的图象经过点A，B，

1bc0 ······················································································· 2分 可得方程组93bc0解得：b4················································ 3分 二次函数解析式为yx24x3 ·

c3（2）过点M作MFx轴，垂足为F． ····································································· 4分

OM是O1的切线，M为切点，． O1MOM（圆的切线垂直于经过切点的半径）

在Rt△OO1M中，sinO1OMO1M1 OO12y P P2 1O M B x  ······························ 5分 O1OM为锐角，OOM301OMOO1cos30233， 2H A F O1 cos303在Rt△MOF中，OFOM13MFOMsin303．

2233． 2233································································································ 6分 点M坐标为2，2 ·

 7

设切线OM的函数解析式为ykx(k0)，由题意可知

333 ······ 7分 k，k223切线OM的函数解析式为y3········································································· 8分 x ·3（3）存在． ····················································································································· 9分 ①过点A作AP1x轴，与OM交于点P1∽Rt△MOO1（两角对应1．可得Rt△APO相等两三角形相似）

33，P·············································· 10分 PtanAOP11AOA1tan301，3 ·3H． ②过点A作AP2OM，垂足为P2，过P2点作P2HOA，垂足为

可得Rt△APO∽Rt△O1MO（两角对应相等两三角开相似） 2OA1，OP2OA在Rt△OPcos302A中，

3， 2在Rt△OPcosAOP222H中，OHOP333， 22433313，P2， ········································· 11分 P2HOP2sinAOP244·224符合条件的P点坐标有1，

6、（08山东济宁26题）（12分）



333，， ··························································· 12分 344△ABC中，C90，A60，AC2cm．长为1cm的线段MN在△ABC的

边AB上沿AB方向以1cm/s的速度向点B运动（运动前点M与点A重合）．过M，N分别作AB的垂线交直角边于P，Q两点，线段MN运动的时间为

ts．

（1）若△AMP的面积为y，写出y与t的函数关系式（写出自变量t的取值范围）；

（2）线段MN运动过程中，四边形MNQP有可能成为矩形吗？若有可能，求出此时t的值；若不可能，说明理由；

（3）t为何值时，以C，P，Q为顶点的三角形与△ABC相似？

8

（08山东济宁26题解析）解：（1）当点P在AC上时，AMt，

PMAMtg603t．

132············································································· 2分 yt3tt(0≤t≤1)． ·

22当点P在BC上时，PMBMtan303(4t)． 3133223······················································ 4分 yt(4t)tt(1≤t≤3)． ·

2363（2）AC2，AB4．BNABAMMN4t13t．

QNBNtan303·············································································· 6分 (3t)． ·

33(3t)， 3由条件知，若四边形MNQP为矩形，需PMQN，即3tt3． 43····································································· 8分 当ts时，四边形MNQP为矩形． ·

43（3）由（2）知，当ts时，四边形MNQP为矩形，此时PQ∥AB，

4△PQC∽△ABC． ··································································································· 9分

除此之外，当CPQB30时，△QPC∽△ABC，此时

CQ3tan30． CP3AM1cos60，AP2AM2t．CP22t． ······························· 10分 AP2BN23BN3，BQ(3t)． cos303BQ2322323t(3t)． ············································ 11分 33又BC23，CQ2323t313，t． 22t32 9

当t13s或s时，以C，P，Q为顶点的三角形与△ABC相似． ···················· 12分 24

7、（08四川巴中30题）（12分）30．已知：如图14，抛

32x3与x轴交于点A，点B，与直线433yxb相交于点B，点C，直线yxb与y44轴交于点E．

（1）写出直线BC的解析式． （2）求△ABC的面积．

（3）若点M在线段AB上以每秒1个单位长度的速度从A向B运动（不与A，B重合），同时，点N在射线BC上以每秒2个单位长度的速度从B向C运动．设运动时间为t秒，请写出△MNB的面积S与t的函数关系式，并求出点M运动多少时间时，△MNB的面积最大，最

物线y大面积是多少？

（08四川巴中30题解析）解：（1）在y32x3中，令y0 4y C E N 3x230

4x12，x22

A(2，0)，B(2，0) ··················································· 1分

又点B在y3xb上 4A M D O P B x 30b

23b

233BC的解析式为yx ····················································································· 2分

4232yx3x114（2）由，得9

33y1yx442x22 ························································· 4分 y0290) C1，，B(2，4 10

9 ······································································································· 5分 4199S△ABC4 ································································································· 6分

242（3）过点N作NPMB于点P EOMB NP∥EO

△BNP∽△BEO ········································································································ 7分 BNNP ···················································································································· 8分 BEEOAB4，CD由直线y333x可得：E0， 422在△BEO中，BO2，EO35，则BE 2262tNP，NPt ······························································································· 9分 5352216St(4t)

25312St2t(0t4) ··························································································· 10分

55312S(t2)2 ····································································································· 11分

5512此抛物线开口向下，当t2时，S最大

512····························· 12分 当点M运动2秒时，△MNB的面积达到最大，最大为． ·58、（08新疆自治区24题）（10分）某工厂要赶制一批抗震救灾用的大型活动板房．如图，板房一面的形状是由矩形和抛物线的一部分组成，矩形长为12m，抛物线拱高为5.6m． （1）在如图所示的平面直角坐标系中，求抛物线的表达式． （2）现需在抛物线AOB的区域内安装几扇窗户，窗户的底边在AB上，每扇窗户宽1.5m，高1.6m，相邻窗户之间的间距均为0.8m，左右两边窗户的窗角所在的点到抛物线的水平距离至少为0.8m．请计算最多可安装几扇这样的窗户？

11

（08新疆自治区24题解析）24．（10分）解：（1）设抛物线的表达式为yax2 1分

5.6)在抛物线的图象上． 点B(6，∴5.636a

7 ··································································· 3分 4572x ·∴抛物线的表达式为y··················································································· 4分 45a（2）设窗户上边所在直线交抛物线于C、D两点，D点坐标为（k，t）

已知窗户高1.6m，∴t5.6(1.6)4 ································································ 5分

4

72k 45·················································································· 6分 k1≈5.07，k2≈5.07（舍去） ·

∴CD5.072≈10.14（m） ····················································································· 7分

又设最多可安装n扇窗户

∴1.5n0.8(n1)≤10.14 ····························································································· 9分

n≤4.06．

12

答：最多可安装4扇窗户． ··························································································· 10分 （本题不要求学生画出4个表示窗户的小矩形）

9、（08广东梅州23题）23．本题满分11分．

如图11所示，在梯形ABCD中，已知AB∥CD， AD⊥DB，AD=DC=CB，AB=4．以AB所在直线为x轴，过D且垂直于AB的直线为y轴建立平面直角坐标系．

（1）求∠DAB的度数及A、D、C三点的坐标；

（2）求过A、D、C三点的抛物线的解析式及其对称轴L． （3）若P是抛物线的对称轴L上的点，那么使PDB为等腰三角形的点P有几个?（不必求点P的坐标，只需说明理由）

（08广东梅州23题解答）解： （1） DC∥AB，AD=DC=CB，

····················································································· 0.5分  ∠CDB=∠CBD=∠DBA， ·

∠DAB=∠CBA， ∠DAB=2∠DBA， ·············· 1分

∠DAB+∠DBA=90， ∠DAB=60， ··········· 1.5分 ∠DBA=30，AB=4， DC=AD=2， ·········· 2分 RtAOD，OA=1，OD=3， ····························· 2.5分 ，D（0， 3），C（2， 3）． · 4分 A（-1，0）

（2）根据抛物线和等腰梯形的对称性知，满足条件的抛物

线必过点A（－1，0），B（3，0）， 故可设所求为 y=a （x+1）（ x-3） ··································································· 6分 将点D（0，

3）的坐标代入上式得， a=3． 3所求抛物线的解析式为 y=3············································· 7分 (x1)(x3). ·

3其对称轴L为直线x=1． ····························································································· 8分 （3） PDB为等腰三角形，有以下三种情况：

①因直线L与DB不平行，DB的垂直平分线与L仅有一个交点P1，P1D=P1B， P1DB为等腰三角形； ························································································· 9分 ②因为以D为圆心，DB为半径的圆与直线L有两个交点P2、P3，DB=DP2，DB=DP3， P2DB， P3DB为等腰三角形；

③与②同理，L上也有两个点P4、P5，使得 BD=BP4，BD=BP5． ······················· 10分 由于以上各点互不重合，所以在直线L上，使PDB为等腰三角形的点P有5个．

13

10、（08广东中山22题）将两块大小一样含30°角的直角三角板，叠放在一起，使得它们的斜边

AB重合，直角边不重合，已知AB=8，BC=AD=4，AC与BD相交于点E，连结CD． (1)填空：如图9，AC= ，BD= ；四边形ABCD是 梯形. (2)请写出图9中所有的相似三角形（不含全等三角形）.

(3)如图10，若以AB所在直线为x轴，过点A垂直于AB的直线为y轴建立如图10的

平面直角坐标系，保持ΔABD不动，将ΔABC向x轴的正方向平移到ΔFGH的位置，FH与BD相交于点P，设AF=t，ΔFBP面积为S，求S与t之间的函数关系式，并写出t的取值值范围.

y D E A 图9

B A F 图10

C

D C E P B G x H （08广东中山22题解析）解：（1）43，43，„„„„„„„„„„1分

等腰；„„„„„„„„„„2分

（2）共有9对相似三角形.（写对3－5对得1分，写对6－8对得2分，写对9对得3分）

①△DCE、△ABE与△ACD或△BDC两两相似，分别是：△DCE∽△ABE，△DCE∽△ACD，△DCE∽△BDC，△ABE∽△ACD，△ABE∽△BDC；(有5对)

②△ABD∽△EAD，△ABD∽△EBC；(有2对) ③△BAC∽△EAD，△BAC∽△EBC；(有2对)

所以，一共有9对相似三角形.„„„„„„„„„„„„„„„„5分

y

（3）由题意知，FP∥AE， ∴ ∠1＝∠PFB，

又∵ ∠1＝∠2＝30°，

DCH ∴ ∠PFB＝∠2＝30°,

∴ FP＝BP.„„„„„„„„„„6分 过点P作PK⊥FB于点K，则FKBK∵ AF＝t，AB＝8，

1FB. 21AFEP21∴ FB＝8－t，BK(8t).

2在Rt△BPK中，PKBKtan2K 图10BGx13(8t)tan30(8t). „„„„„„„„7分 26∴ △FBP的面积S113FBPK(8t)(8t)， 226 14

∴ S与t之间的函数关系式为： S332416(t8)2，或Stt3. „„„„„„„„„„„„„8分 121233t的取值范围为：0t8. „„„„„„„„„„„„„„„„„„„„„„9分 11、（08湖北十堰25题）已知抛物线yax22axb与x轴的一个交点为A(-1,0)，与y轴的正半轴交于点C．

⑴直接写出抛物线的对称轴，及抛物线与x轴的另一个交点B的坐标； ⑵当点C在以AB为直径的⊙P上时，求抛物线的解析式；

⑶坐标平面内是否存在点M，使得以点M和⑵中抛物线上的三点A、B、C为顶点的四边形是平行四边形？若存在，请求出点M的坐标；若不存在，请说明理由．

（08湖北十堰25题解析）解：⑴对称轴是直线：x1，点B的坐标是(3,0)． „„2分

说明：每写对1个给1分，“直线”两字没写不扣分．

⑵如图，连接PC，∵点A、B的坐标分别是A(-1,0)、B (3,0)，

11∴AB＝4．∴PCAB42．

22在Rt△POC中，∵OP＝PA－OA＝2－1＝1， ∴OCPC2PO222123．

∴b＝3． „„„„„„„„„„„„3分 当x1，y0时，a2a30，

∴a3． „„„„„„„„„„„„4分 33223xx3． „„„„„„5分 33∴y⑶存在．„„„„„„„„„„„6分

理由：如图，连接AC、BC．设点M的坐标为M(x,y)．

①当以AC或BC为对角线时，点M在x轴上方，此时CM∥AB，且CM＝AB．

15

由⑵知，AB＝4，∴|x|＝4，yOC3．

∴x＝±4．∴点M的坐标为M(4,3)或(4,3)．„9分

说明：少求一个点的坐标扣1分．

②当以AB为对角线时，点M在x轴下方． 过M作MN⊥AB于N，则∠MNB＝∠AOC＝90°．

∵四边形AMBC是平行四边形，∴AC＝MB，且AC∥MB．

∴∠CAO＝∠MBN．∴△AOC≌△BNM．∴BN＝AO＝1，MN＝CO＝3． ∵OB＝3，∴0N＝3－1＝2．

∴点M的坐标为M(2,3)． „„„„„„„„„„„12分

说明：求点M的坐标时，用解直角三角形的方法或用先求直线解析式，

然后求交点M的坐标的方法均可，请参照给分．

综上所述，坐标平面内存在点M，使得以点A、B、C、M为顶点的四边形是平行四边形．其坐标为M1(4,3),M2(4,3),M3(2,3)．

说明：①综上所述不写不扣分；②如果开头“存在”二字没写，但最后解答全部正确，

不扣分。 12、（08四川达州23题）如图，将△AOB置于平面直角坐标系中，其中点O为坐标原点，

0)，ABO60． 点A的坐标为(3，（1）若△AOB的外接圆与y轴交于点D，求D点坐标．

，0)，试猜想过D，C的直线与△AOB的外接圆的位置关系，并（2）若点C的坐标为(1y

B D 16

加以说明．

（3）二次函数的图象经过点O和A且顶点在圆上， 求此函数的解析式．

0

（08四川达州23题解析）解：（1）连结AD，则∠ADO＝∠B＝60

0

在Rt△ADO中，∠ADO＝60 所以OD＝OA÷3＝3÷3＝3 所以D点的坐标是（0，3）

（2）猜想是CD与圆相切

∵ ∠AOD是直角，所以AD是圆的直径

又∵ Tan∠CDO=CO/OD=1/3=3, ∠CDO＝30

0

F E

F D y B ∴∠CDA=∠CDO+∠ADO=Rt∠ 即CD⊥AD ∴ CD切外接圆于点D

（3）依题意可设二次函数的解析式为 ：

y=α（x－0）(x－3)

由此得顶点坐标的横坐标为：x=E C O A x 3a3=; 2a210

∠B＝30 2即顶点在OA的垂直平分线上，作OA的垂直平分线EF，则得∠EFA＝

3333 可得一个顶点坐标为（，3）

222313） 同理可得另一个顶点坐标为（，22得到EF＝3EA＝

分别将两顶点代入y=α（x－0）(x－3)可解得α的值分别为2323， 39则得到二次函数的解析式是y=2323x(x－3)或y= x(x－3) 3913、（08湖北仙桃等4市25题）如图，直角梯形OABC中，AB∥OC,O为坐标原点，点A在y轴正半轴上，点C在x轴正半轴上，点B坐标为（2，23），∠BCO= 60°，

OHBC于点H.动点P从点H出发，沿线段HO向点O运动，动点Q从点O出发，沿

线段OA向点A运动，两点同时出发，速度都为每秒1个单位长度.设点P运动的时间为t秒.

（1） 求OH的长；

17

（2） 若OPQ的面积为S（平方单位）. 求S与t之间的函数关系式.并求t为何

值时，OPQ的面积最大，最大值是多少？

（3） 设PQ与OB交于点M.①当△OPM为等腰三角形时，求（2）中S的值. ②探究线段OM长度的最大值是多少，直接写出结论. y

B A

H Q M

P

O x C

（08湖北仙桃等4市25题解析）解：（1）∵AB∥OC ∴ OABAOC90 在RtOAB中，AB2 ,AO23

0 ∴OB4， ABO60

0y A B M H P ∴BOC60 而BCO60

∴BOC为等边三角形 ∴OHOBcos30400Q O 323„（3分） 2（2）∵OPOHPH23t

0C x t3t ypOPsin3003

22113∴SOQxpt(3t)

222323=tt （0t23）„„„„„„„„„„（6分）

42333即S (t3)24433∴当t3时，S最大„„„„„„„„„„„„„„„（7分） 4y （3）①若OPM为等腰三角形，则：

B A （i）若OMPM，MPOMOPPOC ∴PQ∥OC

tH ∴OQyp 即t3 MQ 2P 23解得：t O C3 323232323()此时S„„„„„„„„„„„„（8分） 43233∴xpOPcos3030x y

A B 18 Q MH 0（ii）若OPOM，OPMOMP75

∴OQP450

过P点作PEOA，垂足为E，则有： EQEP 即t(313t)3t 22解得：t2

3322233„„„„„„„„„„„„„„（9分） 42（iii）若OPPM，POMPMOAOB

∴PQ∥OA

此时Q在AB上，不满足题意.„„„„„„„„„„„„„„„„„（10分）

3 ②线段OM长的最大值为„„„„„„„„„„„„„„„„„„„„(12分)

214、（08甘肃兰州28题）（本题满分12分）如图19-1，OABC是一张放在平面直角坐标系

此时S中的矩形纸片，O为原点，点A在x轴的正半轴上，点C在y轴的正半轴上，OA5，

OC4．

（1）在OC边上取一点D，将纸片沿AD翻折，使点O落在BC边上的点E处，求D，E两点的坐标；

（2）如图19-2，若AE上有一动点P（不与A，E重合）自A点沿AE方向向E点匀速运动，运动的速度为每秒1个单位长度，设运动的时间为t秒（0t5），过P点作ED的平行线交AD于点M，过点M作AE的平行线交DE于点N．求四边形PMNE的面积S与时间t之间的函数关系式；当t取何值时，S有最大值？最大值是多少？

（3）在（2）的条件下，当t为何值时，以A，M，E为顶点的三角形为等腰三角形，并求出相应的时刻点M的坐标．

y y E E C C B B N D D P M x x O O A A

图19-1 图19-2

（08甘肃兰州28题解析）（本题满分12分） 解：（1）依题意可知，折痕AD是四边形OAED的对称轴， 在Rt△ABE中，AEAO5，AB4．

BEAE2AB252423．CE2．

E点坐标为（2，4）． ········································································································· 2分

222在Rt△DCE中，DCCEDE， 又DEOD．

(4OD)222OD2 ． 解得：CD

5． 219

5D点坐标为0， ············································································································· 3分

2△APM∽△AED． （2）如图①PM∥ED，PMAP5，又知APt，ED，AE5 EDAE2t5tPM， 又PE5t．

522而显然四边形PMNE为矩形．

t15S矩形PMNEPMPE(5t)t2t ······························································· 5分

222S四边形PMNE51525t，又05

22282525时，S矩形PMNE有最大值． ··············································································· 6分 28（3）（i）若以AE为等腰三角形的底，则MEMA（如图①） 在Rt△AED中，MEMA，PMAE，P为AE的中点，

15tAPAE．

22y 又PM∥ED，M为AD的中点． E C B 过点M作MFOA，垂足为F，则MF是△OAD的中位线， N P D 1515MFOD，OFOA， M 2422当t当t55时，05，△AME为等腰三角形． 22O F 图① A x 此时M点坐标为，． ··································································································· 8分 （ii）若以AE为等腰三角形的腰，则AMAE5（如图②）

55245525． 在Rt△AOD中，ADODAO522222y C N D M O F 图② A x E P B 过点M作MFOA，垂足为F． PM∥ED，△APM∽△AED．

APAM． AEAD1AMAE55tAP25，PMt5．

52AD52MFMP5，OFOAAFOAAP525，

（0255），此时M点坐标为(525，5)． ···························· 11分 当t25时，

20

综合（i）（ii）可知，t5或t25时，以A，M，E为顶点的三角形为等腰三角形，相2应M点的坐标为，或(525，5)． ······································································ 12分 15、（08天津市卷26题）（本小题10分） 已知抛物线y3ax22bxc，

（Ⅰ）若ab1，c1，求该抛物线与x轴公共点的坐标；

（Ⅱ）若ab1，且当1x1时，抛物线与x轴有且只有一个公共点，求c的取值范围； x21时，（Ⅲ）若abc0，且x10时，对应的y10；对应的y20，试判断当0x15524时，抛物线与x轴是否有公共点？若有，请证明你的结论；若没有，阐述理由． （08天津市卷26题解析）解（Ⅰ）当ab1，c1时，抛物线为y3x22x1， 方程3x22x10的两个根为x11，x21． 3∴该抛物线与x轴公共点的坐标是1··················································· 2分 0． ·，0和，（Ⅱ）当ab1时，抛物线为y3x22xc，且与x轴有公共点．

1

3

1对于方程3x22xc0，判别式412c≥0，有c≤． ··········································· 3分

3①当c111时，由方程3x22x0，解得x1x2． 333此时抛物线为y3x22x11与x轴只有一个公共点，··································· 4分 0． ·33②当c1时， 3x11时，y132c1c， x21时，y232c5c．

1由已知1x1时，该抛物线与x轴有且只有一个公共点，考虑其对称轴为x，

3y1≤0，1c≤0，应有 即

y0.5c0.2解得5c≤1．

1综上，c或5c≤1． ······················································································· 6分

3（Ⅲ）对于二次函数y3ax22bxc，

21

由已知x10时，y1c0；x21时，y23a2bc0， 又abc0，∴3a2bc(abc)2ab2ab． 于是2ab0．而bac，∴2aac0，即ac0．

∴ac0． ····················································································································· 7分 ∵关于x的一元二次方程3ax22bxc0的判别式

4b212ac4(ac)212ac4[(ac)2ac]0，

∴抛物线y3ax22bxc与x轴有两个公共点，顶点在x轴下方． ································ 8分 又该抛物线的对称轴xb， 3a由abc0，c0，2ab0， 得2aba， ∴

y 1b2． 33a3O 1 x 又由已知x10时，y10；x21时，y20，观察图象，

可知在0x1范围内，该抛物线与x轴有两个公共点． ················································ 10分 16、（08江苏镇江28题）（本小题满分8分）探索研究 如图，在直角坐标系xOy中，点P为函数y12x在第一象限内的图象上的任一点，点A41)，直线l过B(0，1)且与x轴平行，过P作y轴的平行线分别交x轴，l于的坐标为(0，C，Q，连结AQ交x轴于H，直线PH交y轴于R．

（1）求证：H点为线段AQ的中点； （2）求证：①四边形APQR为平行四边形；

②平行四边形APQR为菱形；

（3）除P点外，直线PH与抛物线yA O B R H y P C Q x l 12x有无其它公共点？并说明理由． 4（08江苏镇江28题解析）（1）法一：由题可知AOCQ1．

AOHQCH90，AHOQHC，

△AOH≌△QCH． ································································································ （1分）

22

OHCH，即H为AQ的中点． ··········································································· （2分）

1)，B(0，法二：A(0，······························································ （1分） 1)，OAOB． ·

又BQ∥x轴，HAHQ． ····················································································· （2分） （2）①由（1）可知AHQH，AHRQHP，

AR∥PQ，RAHPQH，

△RAH≌△PQH． ································································································· （3分） ARPQ，

又AR∥PQ，四边形APQR为平行四边形． ························································ （4分）

1)，则PQ1②设Pm，m，PQ∥y轴，则Q(m，过P作PGy轴，垂足为G，在Rt△APG中，

14212m． 4111APAG2PG2m21m2m21m21PQ．

444····················································································· （6分） 平行四边形APQR为菱形． ·

（3）设直线PR为ykxb，由OHCH，得H22m1，2，Pm，m2代入得： 24mmkb0，k，m1222PRyxm． · 直线为························· （7分） 1241kmbm2.bm2.44设直线PR与抛物线的公共点为x，x，代入直线PR关系式得：

14212m111xxm20，(xm)20，解得xm．得公共点为m，m2． 42444所以直线PH与抛物线y

23

12x只有一个公共点P． ················································ （8分） 4

24