30.(09年贵州黔东南州)26、(12分)已知二次函数。

(1)求证：不论a为何实数，此函数图象与x轴总有两个交点。

(2)设a<0，当此函数图象与x轴的两个交点的距离为时，求出此二次函数的解析式。

(3)若此二次函数图象与x轴交于A、B两点，在函数图象上是否存在点P，使得△PAB的面积为，若存在求出P点坐标，若不存在请说明理由。

(09年贵州黔东南州26题解析)解(1)因为△=

所以不论a为何实数，此函数图象与x轴总有两个交点。…………(2分)

(2)设x₁、x₂是的两个根，则，，因两交点的距离是，所以。…………(4分)

即：

变形为：……………………………………(5分)

所以：

整理得：

解方程得：

又因为：a<0

所以：a=－1

所以：此二次函数的解析式为…………………………(6分)

(3)设点P的坐标为，因为函数图象与x轴的两个交点间的距离等于，所以：AB=……………………………………………………………………(8分)

所以：S_△PAB=

所以：

即：，则…………………………………(10分)

当时，，即

解此方程得：=－2或3

当时，，即

解此方程得：=0或1……………………………………(11分)

综上所述，所以存在这样的P点，P点坐标是(－2,3), (3,3), (0, －3)或(1, －3)。…(12分)

29.(09年广西梧州)26．(本题满分12分)

如图(9)-1，抛物线经过A(，0)，C(3，)两点，与轴交于点D，与轴交于另一点B．

(1)求此抛物线的解析式；

(2)若直线将四边形ABCD面积二等分，求的值；

(3)如图(9)-2，过点E(1，1)作EF⊥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⊥轴于点G，若线段MG︰AG＝1︰2，求点M，N的坐标．

(09年广西梧州26题解析)(1)解：把A(，0)，C(3，)代入抛物线 　得

　　 　················································································· 1分

　 整理得

　 　……………… 2分　 　　　　　解得………………3分

　 ∴抛物线的解析式为 ··································································· 4分

　 (2)令　　 解得

　 ∴ B点坐标为(4，0)

　 又∵D点坐标为(0，)　∴AB∥CD　 ∴四边形ABCD是梯形．

　 ∴S_梯形ABCD ＝···························· 5分

设直线与x轴的交点为H，

与CD的交点为T，

则H(，0)，　 T(，)··················· 6分

∵直线将四边形ABCD面积二等分

∴S_梯形AHTD ＝S_梯形ABCD＝4

∴····································· 7分　

∴···························································· 8分

(3)∵MG⊥轴于点G，线段MG︰AG＝1︰2

　　 ∴设M(m，)，········································ 9分　

∵点M在抛物线上　　 ∴　

解得(舍去) ···························· 10分

∴M点坐标为(3，)····················································································· 11分

根据中心对称图形性质知，MQ∥AF，MQ＝AF，NQ＝EF，

∴N点坐标为(1，) ···················································································· 12分

28.(09年广西钦州)26．(本题满分10分)

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＝－，c＝－3．·························· 3分

(2)由(1)，得y＝x²－x－3，它与x轴交于A，B两点，得B(4，0)．

∴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＝x－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＝．···························································································· 7分

若△PHQ∽△COQ，则PH∶CO＝HQ∶OQ，得＝，

即t²+2t－1＝0．

∴t₁＝－1，t₂＝－－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＝．···························································································· 9分

若△PHQ∽△COQ，则PH∶CO＝HQ∶OQ，得＝，

即t²－2t+1＝0．

∴t₁＝t₂＝1(舍去)．·········································································· 10分

综上所述，存在的值，t₁＝－1，t₂＝，t₃＝．························ 10分

27.(09年广西南宁)26．如图14，要设计一个等腰梯形的花坛，花坛上底长米，下底长米，上下底相距米，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．设甬道的宽为米．

(1)用含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分

(2)依题意：·············································· 4分

整理得：

(不符合题意，舍去)······································································· 6分

甬道的宽为5米．

(3)设建设花坛的总费用为万元．

············································ 7分

当时，的值最小．··························································· 8分

因为根据设计的要求，甬道的宽不能超过6米，

米时，总费用最少．······················································································ 9分

最少费用为：万元·················································· 10分

26.(09年广西柳州)26．(本题满分10分)

如图11，已知抛物线()与轴的一个交点为，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点在抛物线的对称轴上，点在抛物线上，且以四点为顶点的四边形为平行四边形，求点的坐标．

(09年广西柳州26题解析)解：(1)对称轴是直线：，

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作于点M，

解法一：利用

∵点A、D、C的坐标分别是A (3，0)，D(1，)、

C(0，)，

∴AO＝3，MD=1．

由得

∴ ··························································································· 3分

又∵······························································ 4分

∴由 得　 ································································ 5分

∴函数解析式为：　 ······················································ 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，)、C(0，)，

∴，，

∵

∴…①　　 ·············································································· 3分

又∵…②　 ···················································· 4分

由①、②得　　 ·································································· 5分

∴函数解析式为： ··························································· 6分

(3)如图所示，当BAFE为平行四边形时

　　　　　 则∥，并且＝．

　　　　　 ∵=4，∴=4　　　　　　　　　　　　　

由于对称为，

∴点F的横坐标为5．·············································· 7分

将代入得，

∴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(，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，)．　 ································· 10分

综上所述，点F的坐标为(5，12)， (，12)或(1，)．

(其它解法参照给分)

25.(09年广西贺州)28．(本题满分10分)如图，抛物线的顶点为A，与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：．

(3)当最大时，求点P的坐标．　　　　　　

(09年广西贺州28题解析)解：(1)抛物线与y轴的交于点B，

令x=0得y=2．

∴B(0，2) ············································· 1分

　　　　 ∵

∴A(-2，3)··········································· 3分

(2)当点P是 AB的延长线与x轴交点时，

．············································· 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，．

综合上述： ················································································ 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

　　 ∴　 ····································································································· 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．

2．第19题至第25题的其它解法，只要思路清晰，解法正确，都应按步骤给予相应分数．

2．求S₂的值时，还可进行如下变形：

S₂＝ S_△PEF－S_△OEF＝S_△PEF－(S_{四边形PEOF}－S_△PEF)＝2 S_△PEF－S_{四边形PEOF}，再利用第(1)题中的结论．

注意：1．按照评分标准分步评分，不得随意变更给分点；

48.(09年湖北孝感)25．(本题满分12分)

如图，点P是双曲线上一动点，过点P作x轴、y轴的垂线，分别交x轴、y轴于A、B两点，交双曲线y= (0＜k₂＜|k₁|)于E、F两点．

(1)图1中，四边形PEOF的面积S₁=　　 ▲　　 (用含k₁、k₂的式子表示)；(3分)

(2)图2中，设P点坐标为(－4，3)．

①判断EF与AB的位置关系，并证明你的结论；(4分)

②记，S₂是否有最小值？若有，求出其最小值；若没有，请说明理由．(5分)

(09年湖北孝感25题解析)解：(1)；　　　 　　　　　　　　　… ………………………3分

(2)①EF∥AB．　 　　　　　　　　……………………………………4分

证明：如图，由题意可得A(–4，0)，B(0，3)，， ．

∴PA=3，PE=，PB=4，PF=．

∴，

∴．　　 　………………………… 6分

又∵∠APB=∠EPF．

∴△APB ∽△EPF，∴∠PAB=∠PEF．

∴EF∥AB．　　 　　…………………………… 7分

②S₂没有最小值，理由如下：

过E作EM⊥y轴于点M，过F作FN⊥x轴于点N，两线交于点Q．

由上知M(0，)，N(，0)，Q(，)．　　 ……………… 8分

而S_△EFQ= S_△PEF，

∴S₂＝S_△PEF－S_△OEF＝S_△EFQ－S_△OEF＝S_△EOM+S_△FON+S_矩形OMQN

＝

＝

=．　　　　　　　　　　　 ………………………… 10分

当时，S₂的值随k₂的增大而增大，而0＜k₂＜12．　 …………… 11分

∴0＜S₂＜24，s₂没有最小值．　　 　　　　　…………………………… 12分

说明：1．证明AB∥EF时，还可利用以下三种方法．方法一：分别求出经过A、B两点和经过E、F两点的直线解析式，利用这两个解析式中x的系数相等来证明AB∥EF；方法二：利用＝来证明AB∥EF；方法三：连接AF、BE，利用S_△AEF＝S_△BFE得到点A、点B到直线EF的距离相等，再由A、B两点在直线EF同侧可得到AB∥EF．

47.(09年湖北襄樊)26．(本小题满分13分)

如图13，在梯形中，点是的中点，是等边三角形．

　 (1)求证：梯形是等腰梯形；

　 (2)动点、分别在线段和上运动，且保持不变．设求与的函数关系式；

　 (3)在(2)中：①当动点、运动到何处时，以点、和点、、、中的两个点为顶点的四边形是平行四边形？并指出符合条件的平行四边形的个数；

②当取最小值时，判断的形状，并说明理由．

(09年湖北襄樊26题解析)(1)证明：∵是等边三角形

∴·········· 1分

∵是中点

∴

∵

∴

∴·························· 2分

∴

∴梯形是等腰梯形．································································· 3分

(2)解：在等边中，

∴

∴······················································································· 4分

∴ ∴······························································ 5分

∵ ∴·········································· 6分

∴　 ∴··································································· 7分

(3)解：①当时，则有

则四边形和四边形均为平行四边形

∴··························································· 8分

当时，则有

则四边形和四边形均为平行四边形

∴······························································ 9分

∴当或时，以P、M和A、B、C、 D中的两个点为顶点的四边形是平行四边形．

此时平行四边形有4个．····································································· 10分

②为直角三角形····································································· 11分

∵

∴当取最小值时，························································ 12分

∴是的中点，而

∴∴······················································ 13分