中考数学压轴题及答案40例第2部分

中考数学压轴题及答案40例（2）5.如图，在直角坐标系中，点为函数在第一象限内的图象上的任一点，点的坐标为，直线过且与轴平行，过作轴的平行线分别交轴，于，连结交轴于，直线交轴于．（1）求证：点为线段的中点；（2）求证：①四边形为平行四边形；②平行四边形为菱形；（3）除点外，直线与抛物线有无其它公共点？并说明理由．（08江苏镇江28题解析）（1）法一：由题可知．，，．··································································（1分），即为的中点．·······················································（2分）法二：，，．························································（1分）又轴，．····························································（2分）（2）①由（1）可知，，，，．··································································（3分），又，四边形为平行四边形．·············································（4分）②设，轴，则，则．过作轴，垂足为，在中，．平行四边形为菱形．···················································（6分）（3）设直线为，由，得，代入得：  直线为．···························································（7分）设直线与抛物线的公共点为，代入直线关系式得：，，解得．得公共点为．所以直线与抛物线只有一个公共点．······································（8分）6.如图13，已知抛物线经过原点O和x轴上另一点A,它的对称轴x=2 与x轴交于点C，直线y=-2x-1经过抛物线上一点B(-2,m)，且与y轴、直线x=2分别交于点D、E.（1）求m的值及该抛物线对应的函数关系式；（2）求证：① CB=CE ；② D是BE的中点；（3）若P(x，y)是该抛物线上的一个动点，是否存在这样的点P,使得PB=PE,若存在，试求出所有符合条件的点P的坐标；若不存在，请说明理由.（1）∵ 点B(-2,m)在直线y=-2x-1上，∴ m=-2×(-2)-1=3.                    ………………………………（2分）∴ B(-2,3)∵ 抛物线经过原点O和点A，对称轴为x=2，∴ 点A的坐标为(4,0) .               设所求的抛物线对应函数关系式为y=a(x-0)(x-4).  ……………………（3分）将点B(-2,3)代入上式，得3=a(-2-0)(-2-4)，∴ .∴ 所求的抛物线对应的函数关系式为，即. （6分）   （2）①直线y=-2x-1与y轴、直线x=2的交点坐标分别为D(0,-1) E(2,-5).          过点B作BG∥x轴，与y轴交于F、直线x=2交于G，          则BG⊥直线x=2，BG=4.     在Rt△BGC中，BC=.∵ CE=5，∴ CB=CE=5.  ……………………（9分）②过点E作EH∥x轴，交y轴于H，则点H的坐标为H(0,-5).又点F、D的坐标为F(0,3)、D(0,-1)，∴ FD=DH=4，BF=EH=2，∠BFD=∠EHD=90°.          ∴ △DFB≌△DHE （SAS），∴ BD=DE.即D是BE的中点.                   ………………………………（11分）   （3）  存在.                               ………………………………（12分）          由于PB=PE，∴ 点P在直线CD上，∴ 符合条件的点P是直线CD与该抛物线的交点.          设直线CD对应的函数关系式为y=kx+b.          将D(0,-1) C(2,0)代入，得. 解得  .          ∴ 直线CD对应的函数关系式为y=x-1.∵ 动点P的坐标为(x，)，∴ x-1=.                    ………………………………（13分）解得 ，.    ∴ ，.∴ 符合条件的点P的坐标为(，)或(，).…（14分）(注：用其它方法求解参照以上标准给分.)7.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：                      （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 8.已知：如图14，抛物线与轴交于点，点，与直线相交于点，点，直线与轴交于点．（1）写出直线的解析式．（2）求的面积．（3）若点在线段上以每秒1个单位长度的速度从向运动（不与重合），同时，点在射线上以每秒2个单位长度的速度从向运动．设运动时间为秒，请写出的面积与的函数关系式，并求出点运动多少时间时，的面积最大，最大面积是多少？（解析）解：（1）在中，令，，·······································1分又点在上的解析式为························································2分（2）由，得  ······················································4分，，·······························································5分·································································6分（3）过点作于点·································································7分·································································8分由直线可得：在中，，，则，·······························································9分································································10分································································11分此抛物线开口向下，当时，当点运动2秒时，的面积达到最大，最大为．····························12分