2009年全国中考数学压轴题精选精析（三）
25.（09年广西贺州）28．(本题满分10分)如图，抛物线y??y 轴交于点B．
（1）求点A、点B的坐标．
（2）若点P是x轴上任意一点，求证：PA?PB≤AB． （3）当PA?PB最大时，求点P的坐标．
（09年广西贺州28题解析）解：（1）抛物线y??令x=0得y=2．
∴B（0，2） ················································ 1分
11 ∵y??x2?x?2??(x?2)2?3
44∴A（—2，3） ··············································· 3分
（2）当点P是 AB的延长线与x轴交点时，
A y 12x?x?2的顶点为A，与4· B O x 第28题图
12x?x?2与y轴的交于点B， y 4· A B H O P x PA?PB?AB． ················································· 5分
当点P在x轴上又异于AB的延长线与x轴的交点时， 在点P、A、B构成的三角形中，PA?PB?AB．
综合上述：PA?PB≤AB ······················································································· 7分 （3）作直线AB交x轴于点P，由（2）可知：当PA—PB最大时，点P是所求的点 ····· 8分
作AH⊥OP于H．
∵BO⊥OP，
∴△BOP∽△AHP AHHP ∴ ·············································································································· 9分 ?BOOP由（1）可知：AH=3、OH=2、OB=2，
∴OP=4，故P（4，0） ·························································································· 10分 注：求出AB所在直线解析式后再求其与x轴交点P（4，0）等各种方法只要正确也相应给分．
26.（09年广西柳州）26．（本题满分10分）
2，0)，与y轴如图11，已知抛物线y?ax?2ax?b（a?0）与x轴的一个交点为B(?1第28题图
的负半轴交于点C，顶点为D．
（1）直接写出抛物线的对称轴，及抛物线与x轴的另一个交点A的坐标； （2）以AD为直径的圆经过点C． ①求抛物线的解析式；
②点E在抛物线的对称轴上，点F在抛物线上，且以B，A，F，E四点为顶点的四边形为平行四边形，求点F的坐标．
（09年广西柳州26题解析）解：（1）对称轴是直线：x?1， 点A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） （2）如图11，连接AC、AD，过D作DM?y 轴于点M， 解法一：利用△AOC∽△CMD
∵点A、D、C的坐标分别是A （3，0），D（1，?a?b）、 C（0，?b），
∴AO＝3，MD=1．
B C D 图11 O A x y AOOC3b?得? CMMDa1∴3?ab?0 ··································································································· 3分
由
又∵0?a?(?1)?2a?(?1)?b ···································································· 4分
2?a?1?3?ab?0∴由? 得? ······································································ 5分
b?33a?b?0??∴函数解析式为：y?x?2x?3 ·························································· 6分 解法二：利用以AD为直径的圆经过点C
∵点A、D的坐标分别是A （3，0） 、D（1，?a?b）、C（0，?b），
∴AC?9?b2，CD?1?a2，AD?∵AC?CD?AD
∴3?ab?0…① ····················································································· 3分
22224?(?a?b)2
又∵0?a?(?1)2?2a?(?1)?b…② ························································· 4分 由①、②得a?1，b?3 ········································································ 5分 ∴函数解析式为：y?x2?2x?3 ································································ 6分
（3）如图所示，当BAFE为平行四边形时
则BA∥EF，并且BA＝EF．
∵BA=4，∴EF=4
由于对称为x?1，
∴点F的横坐标为5． ················································· 7分
将x?5代入y?x2?2x?3得y?12，
∴F（5，12）． ···························································· 8分 根据抛物线的对称性可知，在对称轴的左侧抛物线上也存在点
B O A F，使得四边形BAEF是平行四边形，此时点F坐标为（?3，12）． ····································································································· 9分 C D 当四边形BEAF是平行四边形时，点F即为点D，
此时点F的坐标为（1，?4）． ···································· 10分
图11
综上所述，点F的坐标为（5，12）， （?3，12）或（1，?4）． （其它解法参照给分）
27.（09年广西南宁）26．如图14，要设计一个等腰梯形的花坛，花坛上底长120米，下底长180米，上下底相距80米，在两腰中点连线（虚线）处有一条横向甬道，上下底之间有两条纵向甬道，各甬道的宽度相等．设甬道的宽为x米． （1）用含x的式子表示横向甬道的面积；
（2）当三条甬道的面积是梯形面积的八分之一时，求甬道的宽；
（3）根据设计的要求，甬道的宽不能超过6米.如果修建甬道的总费用（万元）与甬道的宽度成正比例关系，比例系数是5.7，花坛其余部分的绿化费用为每平方米0.02万元，那么当甬道的宽度为多少米时，所建花坛的总费用最少？最少费用是多少万元？ 图14
（09年广西南宁26题解析）解：（1）横向甬道的面积为：（2）依题意：2?80x?150x?2x?整理得：x?155x?750?0
············································································· 6分 x1?5，x2?150（不符合题意，舍去） ·
22E y F x 120?180x?150x?m2? ···· 2分 21120?180??80 ·················································· 4分 82?甬道的宽为5米．
（3）设建设花坛的总费用为y万元．
?120?180?··············································· 7分 y?0.02???80??160x?150x?2x2???5.7x ·
2???0.04x2?0.5x?240
当x??b0.5??6.25时，y的值最小． ································································ 8分 2a2?0.04因为根据设计的要求，甬道的宽不能超过6米，
?当x?6米时，总费用最少． ····························································································· 9分
最少费用为：0.04?6?0.5?6?240?238.44万元 ······················································· 10分 28.（09年广西钦州）26．（本题满分10分）
3如图，已知抛物线y＝x2＋bx＋c与坐标轴交于A、B、C三点， A点的坐标
43为（－1，0），过点C的直线y＝x－3与x轴交于点Q，点P是线段BC上的一个动点，
4t过P作PH⊥OB于点H．若PB＝5t，且0＜t＜1．
（1）填空：点C的坐标是_▲_，b＝_▲_，c＝_▲_； （2）求线段QH的长（用含t的式子表示）；
（3）依点P的变化，是否存在t的值，使以P、H、Q为顶点的三角形与△COQ相似？若存在，求出所有t的值；若不存在，说明理由．
（09年广西钦州26题解析）解：（1）（0，－3），b＝－
（2）由（1），得y＝
2yQHPAOBxC9，c＝－3． ······························ 3分 4329x－x－3，它与x轴交于A，B两点，得B（4，0）． 44y∴OB＝4，又∵OC＝3，∴BC＝5． Q由题意，得△BHP∽△BOC， HAOBx∵OC∶OB∶BC＝3∶4∶5，
P∴HP∶HB∶BP＝3∶4∶5，
∵PB＝5t，∴HB＝4t，HP＝3t．
C∴OH＝OB－HB＝4－4t．
3由y＝x－3与x轴交于点Q，得Q（4t，0）．
4t∴OQ＝4t． ·································································································· 4分 ①当H在Q、B之间时， QH＝OH－OQ
＝（4－4t）－4t＝4－8t． ······································································ 5分 ②当H在O、Q之间时， QH＝OQ－OH
＝4t－（4－4t）＝8t－4． ······································································ 6分 综合①，②得QH＝｜4－8t｜； ····································································· 6分 （3）存在t的值，使以P、H、Q为顶点的三角形与△COQ相似． ···················· 7分
①当H在Q、B之间时，QH＝4－8t，
4?8t3t若△QHP∽△COQ，则QH∶CO＝HP∶OQ，得＝，
34t7∴t＝． ···································································································· 7分
323t4?8t若△PHQ∽△COQ，则PH∶CO＝HQ∶OQ，得＝，
4t3即t2＋2t－1＝0．
∴t1＝2－1，t2＝－2－1（舍去）． ····················································· 8分 ②当H在O、Q之间时，QH＝8t－4．
8t?43t若△QHP∽△COQ，则QH∶CO＝HP∶OQ，得＝，
34t25∴t＝． ···································································································· 9分
323t8t?4若△PHQ∽△COQ，则PH∶CO＝HQ∶OQ，得＝，
4t32
即t－2t＋1＝0． ∴t1＝t2＝1（舍去）． ················································································· 10分
725综上所述，存在t的值，t1＝2－1，t2＝，t3＝． ··························· 10分
323229.（09年广西梧州）26．（本题满分12分）
[来源学。科。网Z。X。X。K]如图（9）-1，抛物线y?ax?3ax?b经过A（?1，0），C（3，?2）两点，与y轴交于点D，与x轴交于另一点B． （1）求此抛物线的解析式；
（2）若直线y?kx?1(k?0)将四边形ABCD面积二等分，求k的值；
（3）如图（9）-2，过点E（1，1）作EF⊥x轴于点F，将△AEF绕平面内某点旋转180°得△MNQ（点M、N、Q分别与点A、E、F对应），使点M、N在抛物线上，作MG⊥x轴于点G，若线段MG︰AG＝1︰2，求点M，N的坐标．
y y E G A D O C B 2x A O F Q N M B x 图（9）-1
y=kx+1
图（9）-2
2?2）（09年广西梧州26题解析）（1）解：把A（?1，0），C（3，代入抛物线 y?ax?3ax?b
压轴题3
