Question

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）

Answer 1

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