Question

如圖，在平面直角坐標(biāo)系中，拋物線=－++經(jīng)過(guò)A（0，－4）、B（，0）、 C（，0）三點(diǎn)，且-=5．

（1）求、的值；（4分）

（2）在拋物線上求一點(diǎn)D，使得四邊形BDCE是以BC為對(duì)     角線的菱形；（3分）

（3）在拋物線上是否存在一點(diǎn)P，使得四邊形BPOH是以OB為對(duì)角線的菱形？若存在，求出點(diǎn)P的坐標(biāo)，并判斷這個(gè)菱形是否為正方形？若不存在，請(qǐng)說(shuō)明理由．（3分）

Answer 1

解：（1）解法一：

∵拋物線=－++經(jīng)過(guò)點(diǎn)A（0，－4），

  ∴=－4 ……1分

又由題意可知，、是方程－++=0的兩個(gè)根，

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

由已知得（-）=25

又（-）=（+）－4=－24

∴ －24=25                                   

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

當(dāng)=時(shí)，拋物線與軸的交點(diǎn)在軸的正半軸上，不合題意，舍去．

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

解法二：∵、是方程－++c=0的兩個(gè)根，

 即方程2－3+12=0的兩個(gè)根．

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

∴－==5，

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

        （以下與解法一相同．）   

    （2）∵四邊形BDCE是以BC為對(duì)角線的菱形，根據(jù)菱形的性質(zhì)，點(diǎn)D必在拋物線的對(duì)稱軸上，     5分

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

            ∴拋物線的頂點(diǎn)（－，）即為所求的點(diǎn)D．······································· 7分

     （3）∵四邊形BPOH是以OB為對(duì)角線的菱形，點(diǎn)B的坐標(biāo)為（－6，0），

根據(jù)菱形的性質(zhì)，點(diǎn)P必是直線=-3與

拋物線=－--4的交點(diǎn)， ···························································· 8分

        ∴當(dāng)=－3時(shí)，=－×（－3）－×（－3）－4=4，

        ∴在拋物線上存在一點(diǎn)P（－3，4），使得四邊形BPOH為菱形． ·················· 9分

          四邊形BPOH不能成為正方形，因?yàn)槿绻倪呅?i>BPOH為正方形，點(diǎn)P的坐標(biāo)只能是（－3，3），但這一點(diǎn)不在拋物線上．······································································································· 10分