26．如图16.在平面直角坐标系中.直线与轴交于点.与轴交于点.抛物线经过三点． (1)求过三点抛物线的解析式并求出顶点的坐标, (2)在抛物线上是否存在点.使为直角三角形.若存在.直接写出点坐标——青夏教育精英家教网—

84.(08辽宁12市26题)(本题14分)26．如图16，在平面直角坐标系中，直线与轴交于点，与轴交于点，抛物线经过三点．

(1)求过三点抛物线的解析式并求出顶点的坐标；

(2)在抛物线上是否存在点，使为直角三角形，若存在，直接写出点坐标；若不存在，请说明理由；

(3)试探究在直线上是否存在一点，使得的周长最小，若存在，求出点的坐标；若不存在，请说明理由．

(08辽宁12市26题解析)

解：(1)直线与轴交于点，与轴交于点．

，························································································· 1分

点都在抛物线上，

抛物线的解析式为························································ 3分

顶点······························································································· 4分

(2)存在··············································································································· 5分

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

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

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

理由：

解法一：

延长到点，使，连接交直线于点，则点就是所求的点．

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

过点作于点．

点在抛物线上，

在中，，

，，

在中，，

，，··············································· 12分

设直线的解析式为

　解得

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

　解得　

在直线上存在点，使得的周长最小，此时．··· 14分

解法二：

过点作的垂线交轴于点，则点为点关于直线的对称点．连接交于点，则点即为所求．················································································ 11分

过点作轴于点，则，．

，

同方法一可求得．

在中，，，可求得，

为线段的垂直平分线，可证得为等边三角形，

垂直平分．

即点为点关于的对称点．············································· 12分

设直线的解析式为，由题意得

　解得

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

　解得　

在直线上存在点，使得的周长最小，此时．··· 14分

83.(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，矩形的边在轴的负半轴上，边在轴的正半轴上，且，，矩形绕点按顺时针方向旋转后得到矩形．点的对应点为点，点的对应点为点，点的对应点为点，抛物线过点．

(1)判断点是否在轴上，并说明理由；

(2)求抛物线的函数表达式；

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

(08辽宁沈阳26题解析)解：(1)点在轴上················································· 1分

理由如下：

连接，如图所示，在中，，，

，

由题意可知：

点在轴上，点在轴上．········································································· 3分

(2)过点作轴于点

，

在中，，

点在第一象限，

点的坐标为·························································································· 5分

由(1)知，点在轴的正半轴上

点的坐标为

点的坐标为···························································································· 6分

抛物线经过点，

由题意，将，代入中得

　解得

所求抛物线表达式为：······················································· 9分

(3)存在符合条件的点，点．········································································· 10分

理由如下：矩形的面积

以为顶点的平行四边形面积为．

由题意可知为此平行四边形一边，

又

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

依题意设点的坐标为

点在抛物线上

解得，，

，

以为顶点的四边形是平行四边形，

，，

当点的坐标为时，

点的坐标分别为，；

当点的坐标为时，

点的坐标分别为，．··············································· 14分

82.(08广东肇庆25题)(本小题满分10分)

已知点A(a，)、B(2a，y)、C(3a，y)都在抛物线上.

(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、y、y，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5=0，·············································································· (1分)

得，．·················································································· (2分)

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

(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

=S - - ···················································· (5分)

　　　　 =--···································· (6分)

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

(3)如：． ········································································· (8分)

事实上， =45a²+36a．　　　　　　　　　　　　　　　

　　　　 3()=3[5×(2a)²+12×2a-(5a²+12a)] =45a²+36a．············ (9分)

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

81.(08广东茂名25题)(本题满分10分)

如图，在平面直角坐标系中，抛物线=－++经过A(0，－4)、B(，0)、 C(，0)三点，且-=5．

(1)求、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=－++经过点A(0，－4)，

　 ∴=－4 ……1分

又由题意可知，、是方程－++=0的两个根，

∴+=，　 =－=6··································································· 2分

由已知得(-)=25

又(-)=(+)－4=－24

∴ －24=25　　　　　　　　　　　　　　　　　　

解得=± ··········································································································· 3分

当=时，抛物线与轴的交点在轴的正半轴上，不合题意，舍去．

∴=－． ·········································································································· 4分

解法二：∵、是方程－++c=0的两个根，

　即方程2－3+12=0的两个根．

∴=，··········································································· 2分

∴－==5，

　　　　解得 =±······························································································· 3分

　　　　 (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

　　　　　又∵=－－－4=－(+)+　 ································· 6分

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

　　 (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=-3与

抛物线=－--4的交点， ···························································· 8分

　　　　 ∴当=－3时，=－×(－3)－×(－3)－4=4，

　　　　 ∴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

　　有理数的加减，打破了小学数学中的加与减的严格界限，把加、减统一成加法。这都是由于引进了负数，也正是由于引进了负数，小学时我们所熟悉的许多结论在有理数范围内都不一定成立了。下面的几个问题作为本文的结尾，请同学们认真思考并做出回答：

　　 (1)“两个数相加，和一定大于或等于各个加数”吗？

　　 (2)“两个数相减，差一定小于或等于被减数”吗？

　　 (3)“一个数的3倍一定大于这个数的2倍”吗？

　　在初次进行有理数的加减运算时，首先要分清“+”、“－”号是运算符号还是性质符号。刚开始时，最好把性质符号用括号括起来，使性质符号与运算符号分开。如：正2加上负3，应写作，不能写成“”。其次，要牢记运算的法则。第三，减法统一变加法。因为学了相反数后，减去一个数，等于加上这个数的相反数。这是有理数的减法法则，它把减法变成了加法。