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
∵y??14x?x?2??214x?x?2的顶点为A，与
2y · A B O x 第28题图
12x?x?2与y轴的交于点B， y 4· A B 14(x?2)?3
2∴A（—2，3） ··········································· 3分
（2）当点P是 AB的延长线与x轴交点时，
············································ 5分 PA?PB?AB． ·
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．
H O P x 第28题图
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 ∴
AHBO?HPOP ···································································································· 9分
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） 如图11，已知抛物线y?ax2?2ax?b（a?0）与x轴的一个交点为B(?1，0)，与y轴
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．
由
AOCM?OCMDy B C O A x D 图11
得
3a?b1
∴3?ab?0 ·························································································· 3分 又∵0?a?(?1)?2a?(?1)?b ······························································ 4分
?3?ab?0?a?1∴由? 得? ································································ 5分
b?33a?b?0??2∴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?∵AC229?b，CD?221?a，AD?24?(?a?b)2
?CD?AD
2∴3?ab?0…① ·············································································· 3分
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F，使得四边形BAEF是平行四边形，此时点F坐标为（?3，12）． ······························································································9分
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 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）
y E F B C O A D x 图11
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？
（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
x1?5，x2?150（不符合题意，舍去） ······································································· 6分
22图14
120?1802x?150x?m2·· 2分 ? ·
18?120?1802?80 ············································· 4分
?甬道的宽为5米．
（3）设建设花坛的总费用为y万元．
?120?1802?··········································· 7分 y?0.02???80??160x?150x?2x???5.7x·
2???0.04x?0.5x?240
2当x??b2a?0.52?0.04·························································· 8分 ?6.25时，y的值最小．·
因为根据设计的要求，甬道的宽不能超过6米，
···················································································· 9分 ?当x?6米时，总费用最少． ·
最少费用为：0.04?62?0.5?6?240?238.44万元···················································10分 28.（09年广西钦州）26．（本题满分10分）
如图，已知抛物线y＝
34x2＋bx＋c与坐标轴交于A、B、C三点， A点的坐标
34t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，
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＝
3494C AOyQHBPx，c＝－3． ··························· 3分
x2－
94x－3，它与x轴交于A，B两点，得B（4，0）．
AO∴OB＝4，又∵OC＝3，∴BC＝5．
由题意，得△BHP∽△BOC， ∵OC∶OB∶BC＝3∶4∶5， ∴HP∶HB∶BP＝3∶4∶5， ∵PB＝5t，∴HB＝4t，HP＝3t． ∴OH＝OB－HB＝4－4t． 由y＝
34tyQHBPxCx－3与x轴交于点Q，得Q（4t，0）．
∴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，
若△QHP∽△COQ，则QH∶CO＝HP∶OQ，得∴t＝
7324?8t3＝
3t4t，
． ··························································································· 7分
3t3若△PHQ∽△COQ，则PH∶CO＝HQ∶OQ，得即t2＋2t－1＝0．
＝
4?8t4t，
∴t1＝2－1，t2＝－2－1（舍去）． ················································ 8分
②当H在O、Q之间时，QH＝8t－4．
若△QHP∽△COQ，则QH∶CO＝HP∶OQ，得∴t＝
25328t?43＝
3t4t，
． ··························································································· 9分
3t3若△PHQ∽△COQ，则PH∶CO＝HQ∶OQ，得＝
8t?44t，
即t2－2t＋1＝0．
∴t1＝t2＝1（舍去）．···········································································10分 综上所述，存在t的值，t1＝2－1，t2＝
29.（09年广西梧州）26．（本题满分12分）
如图（9）-1，抛物线y?ax2?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 732，t3＝
2532．·························10分
x A O F Q N M B x 图（9）-1
y=kx+1
图（9）-2
2（09年广西梧州26题解析）（1）解：把A（?1，0），C（3，?2）代入抛物线 y?ax?3ax?b