摘要：探索研究 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． (1)法一:由题可知． .. ．························································································· .即为的中点．····································································· 法二:..．·························································· 又轴.．··············································································· 可知.. .. ．·························································································· . 又.四边形为平行四边形．···················································· ②设.轴.则.则． 过作轴.垂足为.在中. ． 平行四边形为菱形．··············································································· (3)设直线为.由.得.代入得: 直线为．························ 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．············································· 80 如图.在平面直角坐标系中.点.点分别在轴.轴的正半轴上.且满足． (1)求点.点的坐标． (2)若点从点出发.以每秒1个单位的速度沿射线运动.连结．设的面积为.点的运动时间为秒.求与的函数关系式.并写出自变量的取值范围． 的条件下.是否存在点.使以点为顶点的三角形与相似?若存在.请直接写出点的坐标,若不存在.请说明理由． (08黑龙江齐齐哈尔28题解析)解:(1) .··················································································· . 点.点分别在轴.轴的正半轴上 ······························································································· (2)求得························································································· (每个解析式各1分.两个取值范围共1分)························································· (3),,, ···························································································································· 81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. ------------ ∴ B ∵ 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). -------- 将点B代入上式.得3=a.∴ . ∴ 所求的抛物线对应的函数关系式为.即. (2)①直线y=-2x-1与y轴.直线x=2的交点坐标分别为D E. 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. -------- ②过点E作EH∥x轴.交y轴于H. 则点H的坐标为H. 又点F.D的坐标为F(0,3).D. ∴ FD=DH=4.BF=EH=2.∠BFD=∠EHD=90°. ∴ △DFB≌△DHE (SAS). ∴ BD=DE. 即D是BE的中点. ------------ (3) 存在. ------------ 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 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=. ------------ 解得 .. ∴ .. ∴ 符合条件的点P的坐标为(.)或(.).-

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