中考数学压轴题精选(二次函数)(16题)和详细解答

11、（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y=－32x2+bx+c经过A（0，－4）、B（x1，0）、C（x2，0）三点，且x2-x1=5．（1）求b、c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y=－32x2+bx+c经过点A（0，－4），∴c=－4……1分又由题意可知，x1、x2是方程－32x2+bx+c=0的两个根，∴x1+x2=23b，x1x2=－23c=6··························································2分由已知得（x2-x1）2=25又（x2-x1）2=（x2+x1）2－4x1x2=49b2－24∴49b2－24=25解得b=±314····························································································3分当b=314时，抛物线与x轴的交点在x轴的正半轴上，不合题意，舍去．∴b=－314．···························································································4分解法二：∵x1、x2是方程－32x2+bx+c=0的两个根，即方程2x2－3bx+12=0的两个根．∴x=4969b32b，·································································2分（第25题图）AxyBCO2∴x2－x1=2969b2=5，解得b=±314··················································································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∵y=－32x2－314x－4=－32（x+27）2+625·····························6分∴抛物线的顶点（－27，625）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x=-3与抛物线y=－32x2-314x-4的交点，····················································8分∴当x=－3时，y=－32×（－3）2－314×（－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2、（08广东肇庆25题）（本小题满分10分）已知点A（a，1y）、B（2a，y2）、C（3a，y3）都在抛物线xxy1252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1y、y2、y3，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5xx122=0，···································································（1分）得01x，5122x．·······································································（2分）∴抛物线与x轴的交点坐标为（0，0）、（512，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ABCS=SADFC梯形-ADEBS梯形-BEFCS梯形·············································（5分）3=22)8117(-21)4417(-21)8144(·······························（6分）=5（个单位面积）······························································（7分）（3）如：)(3123yyy．·······························································（8分）事实上，)3(12)3(523aay=45a2+36a．3（12yy）=3[5×（2a）2+12×2a-（5a2+12a）]=45a2+36a．···········（9分）∴)(3123yyy．········································································（10分）3、（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ABOC的边BO在x轴的负半轴上，边OC在y轴的正半轴上，且1AB，3OB，矩形ABOC绕点O按顺时针方向旋转60后得到矩形EFOD．点A的对应点为点E，点B的对应点为点F，点C的对应点为点D，抛物线2yaxbxc过点AED，，．（1）判断点E是否在y轴上，并说明理由；（2）求抛物线的函数表达式；（3）在x轴的上方是否存在点P，点Q，使以点OBPQ，，，为顶点的平行四边形的面积是矩形ABOC面积的2倍，且点P在抛物线上，若存在，请求出点P，点Q的坐标；若不存在，请说明理由．（08辽宁沈阳26题解析）解：（1）点E在y轴上············································1分理由如下：连接AO，如图所示，在RtABO△中，1AB，3BO，2AO1sin2AOB，30AOB由题意可知：60AOE306090BOEAOBAOE点B在x轴上，点E在y轴上．·······························································3分（2）过点D作DMx轴于点M1OD，30DOMyxO第26题图DECFAB4在RtDOM△中，12DM，32OM点D在第一象限，点D的坐标为3122，·············································································5分由（1）知2EOAO，点E在y轴的正半轴上点E的坐标为(02)，点A的坐标为(31)，···············································································6分抛物线2yaxbxc经过点E，2c由题意，将(31)A，，3122D，代入22yaxbx中得33213312422abab解得89539ab所求抛物线表达式为：2853299yxx················································9分（3）存在符合条件的点P，点Q．······························································10分理由如下：矩形ABOC的面积3ABBO以OBPQ，，，为顶点的平行四边形面积为23．由题意可知OB为此平行四边形一边，又3OBOB边上的高为2····················································································11分依题意设点P的坐标为(2)m，点P在抛物线2853299yxx上28532299mm5解得，10m，2538m1(02)P，，25328P，以OBPQ，，，为顶点的四边形是平行四边形，PQOB∥，3PQOB，当点1P的坐标为(02)，时，点Q的坐标分别为1(32)Q，，2(32)Q，；当点2P的坐标为5328，时，点Q的坐标分别为313328Q，，43328Q，．········································14分4、（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33yx与x轴交于点A，与y轴交于点C，抛物线223(0)3yaxxca经过ABC，，三点．（1）求过ABC，，三点抛物线的解析式并求出顶点F的坐标；（2）在抛物线上是否存在点P，使ABP△为直角三角形，若存在，直接写出P点坐标；若不存在，请说明理由；（3）试探究在直线AC上是否存在一点M，使得MBF△的周长最小，若存在，求出M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33yx与x轴交于点A，与y轴交于点C．(10)A，，(03)C，·············································································1分点AC，都在抛物线上，23033acc333ac抛物线的解析式为2323333yxx·················································3分yxODECFABMAOxyBFC图166顶点4313F，··················································································4分（2）存在································································································5分1(03)P，······························································································7分2(23)P，·····························································································9分（3）存在·····················