2012中考数学压轴题及答案40例(2)
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2012中考数学压轴题及答案40例(2)
5.如图,在直角坐标系xOy 中,点P 为函数2
14
y 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P H 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P H 与抛物线2
14
y 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 ==.
90AO H Q C H ∠=∠=
,AHO QHC ∠=∠,
AOH QCH ∴△≌△. ·
·······················································································(1分) O H C H ∴=,即H 为AQ 的中点. ···································································(2分)
法二:(01)A ,,(01)B -,,O A O B ∴=. ························································(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)Q m -,,则2
1
14P Q m =+.
过P 作PG y ⊥轴,垂足为G ,在R t APG △中,
2
2
2
2
22
22111111444AP AG PG m m m m PQ ⎛⎫⎛⎫
=+=
-+=+=+= ⎪ ⎪
⎝⎭
⎝⎭
. ∴平行四边形APQR 为菱形. ·
············································································(6分) (3)设直线P R 为y kx b =+,由O H C H =,得22m
H ⎛⎫
⎪⎝⎭,,214P m m ⎛⎫ ⎪⎝⎭
,代入得:
2021.4m k b km b m ⎧+=⎪⎪⎨
⎪+=⎪⎩, 22
1.
4
m k b m ⎧
=⎪⎪∴⎨⎪=-⎪⎩,∴直线P R 为2124m y x m =-. ·······················(7分) 设直线P R 与抛物线的公共点为214
x x ⎛⎫
⎪⎝
⎭
,,代入直线P R 关系式得:
2
2
1104
2
4
m x x m -
+
=,
2
1
()04x m -=,解得x m =.得公共点为214m m ⎛⎫ ⎪⎝⎭,. 所以直线P H 与抛物线2
14
y 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),∴ 4
1=
a .
∴ 所求的抛物线对应的函数关系式为)4(4
1-=
x x y
,即x
x y -=
2
4
1. (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 =52
2
=+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⎩⎨
⎧=+-=0
21b k b . 解得
1,
2
1-==
b k .
A B
C
O
D
E
x
y
x =2 G F
H