50.(09年湖南长沙)(本题答案暂缺)26．(本题满分10分)

如图，二次函数()的图象与轴交于两点，与轴相交于点．连结两点的坐标分别为、，且当和时二次函数的函数值相等．

(1)求实数的值；

(2)若点同时从点出发，均以每秒1个单位长度的速度分别沿边运动，其中一个点到达终点时，另一点也随之停止运动．当运动时间为秒时，连结，将沿翻折，点恰好落在边上的处，求的值及点的坐标；

(3)在(2)的条件下，二次函数图象的对称轴上是否存在点，使得以为项点的三角形与相似？如果存在，请求出点的坐标；如果不存在，请说明理由．

49.(09年湖北宜昌)24．已知：直角梯形OABC的四个顶点是O(0，0)，A(，1)， B(s，t)，C(，0)，抛物线y=x²+mx－m的顶点P是直角梯形OABC内部或边上的一个动点，m为常数．

(1)求s与t的值，并在直角坐标系中画出直角梯形OABC；

(2)当抛物线y=x²+mx－m与直角梯形OABC的边AB相交时，求m的取值范围．

　　　　　 　　　　　　　　　　　　　　　　　　　　　　　　　(12分)

(09年湖北宜昌24题解析)(1)如图，在坐标系中标出O，A，C三点，连接OA，OC．

∵∠AOC≠90°， ∴∠ABC=90°，

故BC⊥OC， BC⊥AB，∴B(，1)．(1分，)

即s=，t=1．直角梯形如图所画．(2分)

(大致说清理由即可)

(2)由题意，y=x²+mx－m与 y=1(线段AB)相交，

得，　　　　 (3分)∴1＝x²+mx－m，

由 (x－1)(x+1+m)=0，得．

∵=1<，不合题意，舍去．　　　 (4分)

∴抛物线y=x²+mx-m与AB边只能相交于(，1)，　　

　 ∴≤－m－1≤，∴ ． ①(5分)　

又∵顶点P()是直角梯形OABC的内部和其边上的一个动点，

∴，即　 ． ②　　 (6分)　 　　

∵，

(或者抛物线y=x²+mx－m顶点的纵坐标最大值是1)

∴点P一定在线段AB的下方．　　　　　 (7分)　　　

又∵点P在x轴的上方，

∴，

∴ ． (*8分)

③(9分)　　　

又∵点P在直线y=x的下方，∴，(10分)即

　(*8分处评分后，此处不重复评分)

　　 ④　　　

　 由①②③④ ，得．(12分)　

　　　 说明：解答过程，全部不等式漏写等号的扣1分，个别漏写的酌情处理．

72.(2009年青海)28．矩形在平面直角坐标系中位置如图13所示，两点的坐标分别为，，直线与边相交于点．

(1)求点的坐标；

(2)若抛物线经过点，试确定此抛物线的表达式；

(3)设(2)中的抛物线的对称轴与直线交于点，点为对称轴上一动点，以为顶点的三角形与相似，求符合条件的点的坐标．

(2009年青海26题解析)解：(1)点的坐标为．································ (2分)

(2)抛物线的表达式为．······························································· (4分)

(3)抛物线的对称轴与轴的交点符合条件．

∵，

∴．

∵，

∴．··························· (6分)

∵抛物线的对称轴，

∴点的坐标为．····················································································· (7分)

过点作的垂线交抛物线的对称轴于点．

∵对称轴平行于轴，

∴．

∵，

∴．············································································· (8分)

∴点也符合条件，．

∴，

∴．··············································································· (9分)

∴．

∵点在第一象限，

∴点的坐标为，

∴符合条件的点有两个，分别是，．········································ (11分)

71.(2009年内蒙古呼和浩特)25．(10分)某超市经销一种销售成本为每件40元的商品．据市场调查分析，如果按每件50元销售，一周能售出500件；若销售单价每涨1元，每周销量就减少10件．设销售单价为x元(x≥50)，一周的销售量为y件．

(1)写出y与x的函数关系式(标明x的取值范围)；

(2)设一周的销售利润为S，写出S与x的函数关系式，并确定当单价在什么范围内变化时，利润随着单价的增大而增大？

(3)在超市对该种商品投入不超过10000元的情况下，使得一周销售例如达到8000元，销售单价应定为多少？

(2009年内蒙古呼和浩特25题解析)解：(1)

=································································ 3分

(2)

当时，利润随着单价的增大而增大．······························································ 6分

(3)

················································································································ 8分

当时，成本=不符合要求，舍去．

当时，成本=符合要求．

销售单价应定为80元，才能使得一周销售利润达到8000元的同时，投入不超过10000元．　　 10分

70.(2009年内蒙古包头)26．(本小题满分12分)

已知二次函数()的图象经过点，，，直线()与轴交于点．

(1)求二次函数的解析式；

(2)在直线()上有一点(点在第四象限)，使得为顶点的三角形与以为顶点的三角形相似，求点坐标(用含的代数式表示)；

(3)在(2)成立的条件下，抛物线上是否存在一点，使得四边形为平行四边形？若存在，请求出的值及四边形的面积；若不存在，请说明理由．

(2009年内蒙古包头26题解析)解：(1)根据题意，得

解得．

．······························· (2分)

(2)当时，

得或，

∵，

当时，得，

∴，

∵点在第四象限，∴．································································ (4分)

当时，得，∴，

∵点在第四象限，∴．································································ (6分)

(3)假设抛物线上存在一点，使得四边形为平行四边形，则

，点的横坐标为，

当点的坐标为时，点的坐标为，

∵点在抛物线的图象上，

∴，

∴，

∴，

∴(舍去)，

∴，

∴．························································································· (9分)

当点的坐标为时，点的坐标为，

∵点在抛物线的图象上，

∴，

∴，

∴，∴(舍去)，，

∴，

∴．························································································· (12分)

注：各题的其它解法或证法可参照该评分标准给分．

69.(2009年辽宁铁岭)26．如图所示，已知在直角梯形中，轴于点．动点从点出发，沿轴正方向以每秒1个单位长度的速度移动．过点作垂直于直线，垂足为．设点移动的时间为秒()，与直角梯形重叠部分的面积为．

(1)求经过三点的抛物线解析式；

(2)求与的函数关系式；

(3)将绕着点顺时针旋转，是否存在，使得的顶点或在抛物线上？若存在，直接写出的值；若不存在，请说明理由．

(2009年辽宁铁岭26题解析)26．解：(1)法一：由图象可知：抛物线经过原点，

设抛物线解析式为．

把，代入上式得：···················································································· 1分

解得······················································································· 3分

∴所求抛物线解析式为········································································ 4分

法二：∵，，

∴抛物线的对称轴是直线．

设抛物线解析式为()······························································ 1分

把，代入得

　　 解得··········································································· 3分

∴所求抛物线解析式为．····························································· 4分

(2)分三种情况：

①当，重叠部分的面积是，过点作轴于点，

∵，在中，，，

在中，，，

∴，

∴．··············································· 6分

②当，设交于点，作轴于点，

，则四边形是等腰梯形，

重叠部分的面积是．

∴，

∴．··········· 8分

③当，设与交于点，交于点，重叠部分的面积是．

因为和都是等腰直角三角形，所以重叠部分的面积是．

∵，，

∴，

∴，

∴

　 ．················································ 10分

(3)存在　　 ···································································································· 12分

　　　　　　 ··································································································· 14分

68.(2009年辽宁锦州)26.如图14，抛物线与x轴交于A(x1,0)，B(x2,0)两点，且x1＞x2，与y轴交于点C(0,4)，其中x1,x2是方程x2-2x-8=0的两个根.

　(1)求这条抛物线的解析式；

　(2)点P是线段AB上的动点，过点P作PE∥AC，交BC于点E，连接CP，当△CPE的面积最大时，求点P的坐标；

　(3)探究：若点Q是抛物线对称轴上的点，是否存在这样的点Q，使△QBC成为等腰三角形，若存在，请直接写出所有符合条件的点Q的坐标；若不存在，请说明理由.

(2009年辽宁锦州26题解析)26.解：(1) ∵x2-2x-8=0 ，∴(x-4)(x+2)=0 .∴x1=4,x2=-2.

　∴A(4,0) ,B(-2,0). ……1分

　又∵抛物线经过点A、B、C,设抛物线解析式为y=ax2+bx+c (a≠0)，

　∴ 　∴ ……3分

　∴所求抛物线的解析式为. ……4分

　(2)设P点坐标为(m,0),过点E作EG⊥x轴于点G.

　∵点B坐标为(-2，0)，点A坐标(4,0)，

　∴AB=6， BP=m+2.

　∵PE∥AC，

　∴△BPE∽△BAC.

　∴.

　∴.

　∴S△CPE= S△CBP- S△EBP

　=.

　∴

　 .

　∴. ……7分

　又∵-2≤m≤4，

　∴当m=1时，S△CPE有最大值3.

　此时P点的坐标为(1，0). ……9分

　(3)存在Q点，其坐标为Q1(1，1)，

　，

　，

　，

　.……14分 　

67.(2009年辽宁抚顺)26．已知：如图所示，关于的抛物线与轴交于点、点，与轴交于点．

(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，使四边形为等腰梯形，写出点的坐标，并求出直线的解析式；

(3)在(2)中的直线交抛物线的对称轴于点，抛物线上有一动点，轴上有一动点．是否存在以为顶点的平行四边形？如果存在，请直接写出点的坐标；如果不存在，请说明理由．

(2009年辽宁抚顺26题解析)26．解：(1)根据题意，得

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

解得··············································· 3分

抛物线的解析式为········· 4分

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

(2)················································································································ 6分

设直线的解析式为

直线经过点点

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

······················································································································ 8分

················································································································ 9分

(3)存在．················································································································ 10分

············································································································ 11分

······································································································· 12分

············································································································ 13分

············································································································ 14分

66.(2009年辽宁朝阳)26．如图，点，的坐标分别为(2，0)和(0，)，将绕点按逆时针方向旋转后得，点的对应点是点，点的对应点是点．

(1)写出，两点的坐标，并求出直线的解析式；

(2)将沿着垂直于轴的线段折叠，(点在轴上，点在上，点不与，重合)如图，使点落在轴上，点的对应点为点．设点的坐标为()，与重叠部分的面积为．

i)试求出与之间的函数关系式(包括自变量的取值范围)；

ii)当为何值时，的面积最大？最大值是多少？

iii)是否存在这样的点，使得为直角三角形？若存在，直接写出点的坐标；若不存在，请说明理由．

(2009年辽宁朝阳26题解析)解：(1)···································· (2分)

设直线的解析式，则有

　　 解得

直线的解析式为······································································ (3分)

(2)i)①点在原点和轴正半轴上时，重叠部分是．

则

当与重合时，·················································· (4分)

②当在轴的负半轴上时，设与轴交于点，则重叠部分为梯形.

又

············································ (5分)

当点与点重合时，点的坐标为

············································································································ (6分)

综合得······················································ (7分)

ii)当时，　 对称轴是

抛物线开口向上，在中，随的增大而减小

当时，的最大值＝························································· (8分)

当时，

对称轴是

抛物线开口向下

当时，有最大值为············································································· (9分)

综合当时，有最大值为······························································ (10分)

iii)存在，点的坐标为和···························································· (14分)

附：详解：当以点为直角顶点时，作交轴负半轴于点，

点坐标为(，0)

点的坐标为

当以点为直角顶点时

同样有

点的坐标

综合①②知满足条件的坐标有和．

以上仅提供本试题的一种解法或解题思路，若有不同解法请参照评分标准予以评分．

65.(2009年辽宁本溪)26．如图所示，在平面直角坐标系中，抛物线()经过，，三点，其顶点为，连接，点是线段上一个动点(不与重合)，过点作轴的垂线，垂足为，连接．

(1)求抛物线的解析式，并写出顶点的坐标；

(2)如果点的坐标为，的面积为，求与的函数关系式，写出自变量的取值范围，并求出的最大值；

(3)在(2)的条件下，当取得最大值时，过点作的垂线，垂足为，连接，把沿直线折叠，点的对应点为，请直接写出点坐标，并判断点是否在该抛物线上．

(2009年辽宁本溪26题解析)26．解：(1)设，·························· 1分

把代入，得，························································································· 2分

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

顶点的坐标为．································································································ 5分

(2)设直线解析式为：()，把两点坐标代入，

得·············································································································· 6分

解得．

∴直线解析式为．··············································································· 7分

，······················································· 8分

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

　 ．···························································· 10分

∴当时，取得最大值，最大值为．······························································ 11分

(3)当取得最大值，，，∴．·················································· 5分

∴四边形是矩形．

作点关于直线的对称点，连接．

法一：过作轴于，交轴于点．

设，则．

在中，由勾股定理，

．

解得．

∵，

∴．

由，可得，．

∴．

∴坐标．································································································ 13分

法二：连接，交于点，分别过点作的垂线，垂足为．

易证．

∴．

设，则．

∴，．

由三角形中位线定理，

．

∴，即．

∴坐标．································································································ 13分

把坐标代入抛物线解析式，不成立，所以不在抛物线上．·················· 14分