中考数学压轴题及答案如图,在直角坐标系xOy 中,点P 为函数214y x =在第一象限内的图象上的任一点,点A 的坐标为(01),,直线l 过(01)B -,且与x 轴平行,过P 作y 轴的平行线分别交x 轴,l 于C Q ,,连结AQ 交x 轴于H ,直线PH 交y 轴于R .(1)求证:H 点为线段AQ 的中点; (2)求证:①四边形APQR 为平行四边形;②平行四边形APQR 为菱形;(3)除P 点外,直线PH 与抛物线214y x =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1AO CQ ==.90AOH QCH ∠=∠=,AHO QHC ∠=∠,AOH QCH ∴△≌△. ····················································································· (1分)OH CH ∴=,即H 为AQ 的中点. ································································· (2分) 法二:(01)A ,,(01)B -,,OA OB ∴=. ······················································ (1分) 又BQ x ∥轴,HA HQ ∴=. ·········································································· (2分) (2)①由(1)可知AH QH =,AHR QHP ∠=∠,AR PQ ∥,RAH PQH ∴∠=∠,RAH PQH ∴△≌△. ····················································································· (3分) AR PQ ∴=,又AR PQ ∥,∴四边形APQR 为平行四边形. ··············································· (4分)②设214P m m ⎛⎫ ⎪⎝⎭,,PQ y ∥轴,则(1)Qm -,,则2114PQ m =+.过P 作PG y ⊥轴,垂足为G ,在Rt APG △中,22222222111111444AP AG PG m m m m PQ ⎛⎫⎛⎫=+=-+=+=+= ⎪ ⎪⎝⎭⎝⎭.∴平行四边形APQR 为菱形. ·········································································· (6分)(3)设直线PR 为y kx b =+,由OH CH =,得22m H ⎛⎫⎪⎝⎭,,214P m m ⎛⎫ ⎪⎝⎭,代入得: 2021.4m k b km b m ⎧+=⎪⎪⎨⎪+=⎪⎩, 221.4m k b m ⎧=⎪⎪∴⎨⎪=-⎪⎩,∴直线PR 为2124m y x m =-. ······················ (7分) 设直线PR 与抛物线的公共点为214x x ⎛⎫ ⎪⎝⎭,,代入直线PR 关系式得:22110424m x x m -+=,21()04x m -=,解得x m =.得公共点为214m m ⎛⎫ ⎪⎝⎭,. 所以直线PH 与抛物线214y x =只有一个公共点P . ········································ (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),∴ 41=a . ∴ 所求的抛物线对应的函数关系式为)4(41-=x x y ,即x x y -=241. (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 =522=+BG CG .∵ 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⎩⎨⎧=+-=021b k b . 解得 1,21-==b k . A BCODExy x =2 G FH∴ 直线CD 对应的函数关系式为y =21x -1.∵ 动点P 的坐标为(x ,x x -241),∴21x -1=x x -241. ………………………………(13分) 解得 531+=x ,532-=x . ∴ 2511+=y ,2511-=y . ∴ 符合条件的点P 的坐标为(53+,251+)或(53-,251-).…(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y =-32x 2+b x +c 经过A (0,-4)、B (x 1,0)、 C (x 2,0)三点,且x 2-x 1=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B P O H 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y =-32x 2+b x +c 经过点A (0,-4),∴c =-4 ……1分又由题意可知,x 1、x 2是方程-32x 2+b x +c =0的两个根, ∴x 1+x 2=23b , x 1x 2=-23c =6 ······························································· 2分 由已知得(x 2-x 1)2=25 又(x 2-x 1)2=(x 2+x 1)2-4x 1x 2=49b 2-24 ∴49b 2-24=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:∵x 1、x 2是方程-32x 2+b x +c=0的两个根, 即方程2x 2-3b x +12=0的两个根. ∴x =4969b 32-±b , ······································································· 2分∴x 2-x 1=2969b 2-=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 =-32x 2-314x -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 =-32x 2-314x -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8.已知:如图14,抛物线2334y x =-+与x 轴交于点A ,点B ,与直线34y x b =-+相交于点B ,点C ,直线34y x b =-+与y 轴交于点E . (1)写出直线BC 的解析式. (2)求ABC △的面积.(3)若点M 在线段AB 上以每秒1个单位长度的速度从A 向B 运动(不与A B ,重合),同时,点N 在射线BC 上以每秒2个单位长度的速度从B 向C 运动.设运动时间为t 秒,请写出MNB △的面积S 与t 的函数关系式,并求出点M 运动多少时间时,MNB △的面积最大,最大面积是多少?(解析)解:(1)在2334y x =-+中,令0y = 23304x ∴-+=12x ∴=,22x =-(20)A ∴-,,(20)B ,············································· 1分又点B 在34y x b =-+上 302b ∴=-+32b =BC ∴的解析式为3342y x =-+ ·········································································· 2分 (2)由23343342y x y x ⎧=-+⎪⎪⎨⎪=-+⎪⎩,得11194x y =-⎧⎪⎨=⎪⎩ 2220x y =⎧⎨=⎩················································· 4分 914C ⎛⎫∴- ⎪⎝⎭,,(20)B ,4AB ∴=,94CD =·························································································· 5分 1994242ABC S ∴=⨯⨯=△ ····················································································· 6分 (3)过点N 作NP MB ⊥于点PEO MB ⊥NP EO ∴∥BNP BEO ∴△∽△ ··························································································· 7分 BN NP BE EO∴=····································································································· 8分 由直线3342y x =-+可得:302E ⎛⎫ ⎪⎝⎭, ∴在BEO △中,2BO =,32EO =,则52BE = 25322t NP∴=,65NP t ∴= ··················································································· 9分 16(4)25S t t ∴=- 2312(04)55S t t t =-+<< ················································································ 10分2312(2)55S t =--+ ························································································· 11分 此抛物线开口向下,∴当2t =时,125S =最大 ∴当点M 运动2秒时,MNB △的面积达到最大,最大为125. ······················ 12分。