练习册

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    26．解:(1)由已知.得.. . ． ．············································································································ 设过点的抛物线的解析式为． 将点的坐标代入.得． 将和点的坐标分别代入.得 ··································································································· 解这个方程组.得 故抛物线的解析式为．··························································· (2)成立．························································································· 点在该抛物线上.且它的横坐标为. 点的纵坐标为．······················································································· 设的解析式为. 将点的坐标分别代入.得 解得 的解析式为．········································································ .．··························································································· 过点作于点. 则． . ． 又. ． ． ．··········································································································· ． (3)点在上...则设． ..． ①若.则. 解得．.此时点与点重合． ．··········································································································· ②若.则. 解得 ..此时轴． 与该抛物线在第一象限内的交点的横坐标为1. 点的纵坐标为． ．······································································································· ③若.则. 解得..此时.是等腰直角三角形． 过点作轴于点. 则.设. ． ． 解得． ．··········································· 综上所述.存在三个满足条件的点. 即或或． 26．如图.已知抛物线经过点.抛物线的顶点为.过作射线．过顶点平行于轴的直线交射线于点.在轴正半轴上.连结． (1)求该抛物线的解析式, (2)若动点从点出发.以每秒1个长度单位的速度沿射线运动.设点运动的时间为．问当为何值时.四边形分别为平行四边形?直角梯形?等腰梯形? (3)若.动点和动点分别从点和点同时出发.分别以每秒1个长度单位和2个长度单位的速度沿和运动.当其中一个点停止运动时另一个点也随之停止运动．设它们的运动的时间为.连接.当为何值时.四边形的面积最小?并求出最小值及此时的长． *26．解:(1)抛物线经过点. ·························································································· 1分 二次函数的解析式为:·················································· 3分 (2)为抛物线的顶点过作于.则. ··················································· 4分 当时.四边形是平行四边形 ················································ 5分 当时.四边形是直角梯形 过作于.则 (如果没求出可由求) ····························································································· 6分 当时.四边形是等腰梯形 综上所述:当.5.4时.对应四边形分别是平行四边形.直角梯形.等腰梯形．·· 7分 及已知.是等边三角形 则 过作于.则········································································· 8分 =·································································································· 9分 当时.的面积最小值为··································································· 10分 此时 ······················································ 11分 【查看更多】

     

    题目列表(包括答案和解析)

    已知四边形ABCD中，P是对角线BD上的一点，过P作MN∥AD，EF∥CD，分别交AB、CD、AD、BC于点M、N、E、F，设a=PM•PE，b=PN•PF，解答下列问题：
    （1）当四边形ABCD是矩形时，见图1，请判断a与b的大小关系，并说明理由；
    （2）当四边形ABCD是平行四边形，且∠A为锐角时，见图2，（1）中的结论是否成立？并说明理由；
    （3）在（2）的条件下，设
    BP
    PD
    =k    ，是否存在这样的实数k，使得
    S平行四边形PEAM
    S△ABD
    =
    4
    9
    ？若存在，请求出满足条件的所有k的值；若不存在，请说明理由．
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    已知：在平面直角坐标系xOy中，二次函数y=x2+bx+c的图象与x轴交于A、B两点，点A在点B的左侧，若抛物线的对称轴为x=1，点A的坐标为（-1，0）．
    （1）求这个二次函数的解析式；
    （2）设抛物线的顶点为C，抛物线上一点D的坐标为（-3，12），过点B、D的直线与抛物线的对称轴交于点E．问：是否存在这样的点F，使得以点B、C、E、F为顶点的四边形是平行四边形？若存在，求出点F的坐标；若不存在，请说明理由；
    （3）在（2）的条件下，若在BD上存在一点P，使得直线AP将四边形ACBD分成了面积相等的两部分，请你求出此时点P的坐标．

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    已知：在平面直角坐标系xOy中，二次函数y=x2+bx+c的图象与x轴交于A、B两点，点A在点B的左侧，若抛物线的对称轴为x=1，点A的坐标为（-1，0）．
    （1）求这个二次函数的解析式；
    （2）设抛物线的顶点为C，抛物线上一点D的坐标为（-3，12），过点B、D的直线与抛物线的对称轴交于点E．问：是否存在这样的点F，使得以点B、C、E、F为顶点的四边形是平行四边形？若存在，求出点F的坐标；若不存在，请说明理由；
    （3）在（2）的条件下，若在BD上存在一点P，使得直线AP将四边形ACBD分成了面积相等的两部分，请你求出此时点P的坐标．

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    已知：在平面直角坐标系xOy中，二次函数y=x2+bx+c的图象与x轴交于A、B两点，点A在点B的左侧，若抛物线的对称轴为x=1，点A的坐标为（-1，0）．
    （1）求这个二次函数的解析式；
    （2）设抛物线的顶点为C，抛物线上一点D的坐标为（-3，12），过点B、D的直线与抛物线的对称轴交于点E．问：是否存在这样的点F，使得以点B、C、E、F为顶点的四边形是平行四边形？若存在，求出点F的坐标；若不存在，请说明理由；
    （3）在（2）的条件下，若在BD上存在一点P，使得直线AP将四边形ACBD分成了面积相等的两部分，请你求出此时点P的坐标．

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    已知抛物线y＝x2＋4x＋m(m为常数)经过点(0，4)．

    (1)求m的值；

    (2)将该抛物线先向右、再向下平移得到另一条抛物线．已知平移后的抛物线满足下述两个条件：它的对称轴(设为直线l2)与平移前的抛物线的对称轴(设为直线l1)关于y轴对称；它所对应的函数的最小值为－8．

    ①试求平移后的抛物线的解析式；

    ②试问在平移后的抛物线上是否存在点P，使得以3为半径的圆P既与x轴相切，又与直线l2相交？若存在，请求出点P的坐标，并求出直线l2被圆P所截得的弦AB的长度；若不存在，请说明理由．

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