2010年中考数学压轴题100题精选(81-90题)答案巩固基础
2010年中考数学压轴题100题精选（81-90题）答案
【081】解：（1）（0，－3），b＝－9
4
，c＝－3．···································································· 3分
（2）由（1），得y＝3
4
x2－
9
4
x－3，它与x轴交于A，B两点，得B（4，0）．
∴OB＝4，又∵OC＝3，∴BC＝5．由题意，得△BHP∽△BOC，
∵OC∶OB∶BC＝3∶4∶5，
∴HP∶HB∶BP＝3∶4∶5，
∵PB＝5t，∴HB＝4t，HP＝3t．
∴OH＝OB－HB＝4－4t．
由y＝3
4t
x－3与x轴交于点Q，得Q（4t，0）．
∴OQ＝4t． ··································································
①当H在Q、B之间时，
QH＝OH－OQ
＝（4－4t）－4t＝4－8t． ········································································ 5分
②当H在O、Q之间时，
QH＝OQ－OH
＝4t－（4－4t）＝8t－4． ········································································ 6分综合①，②得QH＝｜4－8t｜； ········································································6分（3）存在t的值，使以P、H、Q为顶点的三角形与△COQ相似． ·······················7分
①当H在Q、B之间时，QH＝4－8t，
若△QHP∽△COQ，则QH∶CO＝HP∶OQ，得48
3
t
-
＝
3
4
t
t
，
∴t＝7
32
． ········································································································7分
若△PHQ∽△COQ，则PH∶CO＝HQ∶OQ，得3
3
t
＝
48
4
t
t
-
，
即t2＋2t－1＝0．
∴t11，t21（舍去）． ·······················································8分②当H在O、Q之间时，QH＝8t－4．
若△QHP∽△COQ，则QH∶CO＝HP∶OQ，得84
3
t-
＝
3
4
t
t
，
∴t＝25
32
． ········································································································9分
若△PHQ∽△COQ，则PH∶CO＝HQ∶OQ，得3
3
t
＝
84
4
t
t
-
，
即t2－2t＋1＝0．
∴t1＝t2＝1（舍去）． ····················································································10分
综上所述，存在t的值，t11，t2＝7
32
，t3＝
25
32
． ···························10分
附加题：解：（1）8；··················································································································5分（2）2．················································································································10分
【082】（09上海）略
【083】.解：（1）B（1
（2）设抛物线的解析式为y =ax (x+a )，代入点B （1,
3，得3
a =， 因此2323y =
+ （3）如图，抛物线的对称轴是直线x =—1，当点C 位于对称轴与线段AB 的交点时，△BOC
的周长最小. 设直线AB 为y =kx +b .所以33,3
20.23
3k k b k b b ???+=????
-+=???=??
解得， 因此直线AB 为323
y =+
， 当x =－1时，3y =
， 因此点C 的坐标为（－13.
（4）如图，过P 作y 轴的平行线交AB 于D .
222
1
()()
2
132332332333
31932PAB PAD PBD D P B A S S S y y x x x x ???=+=--???=+-+?????????????=?=++
???
当x =－
1
2时，△PAB 93，此时13,2P ?- ??
. 【084】解：（1）⊙P 与x 轴相切.
∵直线y =－2x －8与x 轴交于A （4，0），与y 轴交于B （0，－8），
∴OA =4，OB =8.由题意，OP =－k ，∴PB =PA =8+k . 在Rt △AOP 中，k 2+42=(8+k )2， ∴k =－3，∴OP 等于⊙P 的半径， ∴⊙P 与x 轴相切.
（2）设⊙P 与直线l 交于C ，D 两点，连结PC ，PD 当圆心P 在线段OB 上时,作PE ⊥CD 于E .
∵△PCD 为正三角形，∴DE =12CD =3
2
，PD =3， ∴PE 33
. ∵∠AOB =∠PEB =90°， ∠ABO =∠PBE ， ∴△AOB ∽△PEB ，
C
B
A
O
y
x
D
B
A
O
y
x
P
∴
2,
AO PE AB PB PB
=，
∴PB =∴8PO BO PB =-=
∴8)P -，∴8k =.
当圆心P 在线段OB 延长线上时,同理可得P (0,－8)，
∴k =－8，∴当k －8或k =－8时，以⊙P 与直线l 的两个交点和圆心P 为顶点的三角形是正三角形.
【085】解: (1)由题知: ?
?
?=+-=++03390
3b a b a ……………………………………1 分
解得: ??
?-=-=2
1
b a ……………………………………………………………2分
∴ 所求抛物线解析式为: 322+=x －－x y ……………………………3分
(2) 存在符合条件的点P , 其坐标为P (－1, 10)或P(－1,－ 10)
或P (－1, 6) 或P (－1, 3
5
)………………………………………………………7分 (3)解法①：
过点E 作EF ⊥x 轴于点F , 设E ( a ,-2
a -2a ＋3 )( －3< a < 0 ) ∴EF =-2
a -2a ＋3，BF =a ＋3，OF =－a ………………………………………………8 分
∴S 四边形BOCE =
21BF ·EF + 2
1(OC +EF )·OF =21( a ＋3 )·(－2a －2a ＋3) + 21(－2a －2a ＋6)·（－a ）……………………………9 分 =29
29232+--a a ………………………………………………………………………10 分
=－232)2
3(+a +863
∴ 当a =－23
时，S 四边形BOCE 最大, 且最大值为 8
63．……………………………11 分
此时，点E 坐标为 (－23，4
15
)……………………………………………………12分
解法②：
过点E 作EF ⊥x 轴于点F ， 设E ( x , y ) ( －3< x < 0 ) …………………………8分
则S 四边形BOCE =
21(3 + y )·(－x ) + 21( 3 + x )·y ………………………………………9分 = 2
3( y －x )= 23
(332+x －－x ) …………………………………10 分
= －232)2
3
(+x + 863
∴ 当x =－23
时，S 四边形BOCE 最大，且最大值为 8
63． …………………………11分
此时，点E 坐标为 (－23，4
15
) ……………………………………………………12分
【086】⑴证明：∵BC 是⊙O 的直径
∴∠BAC=90o
又∵EM ⊥BC ，BM 平分∠ABC ， ∴AM=ME ，∠AMN=EMN 又∵MN=MN ， ∴△ANM ≌△ENM
⑵∵AB 2=A F ·AC ∴AB
AF AC AB =
又∵∠BAC=∠FAB=90o ∴△ABF ∽△ACB ∴∠ABF=∠C
又∵∠FBC=∠ABC+∠FBA=90o ∴FB 是⊙O 的切线
⑶由⑴得AN=EN ，AM=EM ，∠AMN=EMN ， 又∵AN ∥ME ，∴∠ANM=∠EMN ， ∴∠AMN=∠ANM ，∴AN=AM ， ∴AM=ME=EN=AN ∴四边形AMEN 是菱形 ∵cos ∠ABD=53，∠ADB=90o
∴5
3=AB
BD
设BD=3x ，则AB=5x ，，由勾股定理()()x x －x AD 4352
2==
而AD=12，∴x=3
∴BD=9，AB=15
∵MB 平分∠AME ，∴BE=AB=15 ∴DE=BE-BD=6
∵ND ∥ME ，∴∠BND=∠BME ，又∵∠NBD=∠MBE ∴△BND ∽△BME ，则BE
BD ME ND =
设ME=x ，则ND=12-x ，15912=-x x ，解得x=215
∴S=M E ·DE=2
15×6=45
【087】（天门）略
【088】解：（1）法一：由图象可知：抛物线经过原点， 设抛物线解析式为2
(0)y ax bx a =+≠．
把(11)A ，，(31)B ，代入上式得： ································································································ 1分 11931a b a b =+??
=++?解得1343a b ?
=-????=??
··································································································· 3分 ∴所求抛物线解析式为214
33
y x x =-
+··················································································· 4分 法二：∵(11)
A ，，(31)
B ，， ∴抛物线的对称轴是直线2x =．
设抛物线解析式为2
(2)y a x h =-+（0a ≠） ······································································ 1分
把(00)O ，
，(11)A ，代入得
2
2
0(02)1(12)a h a h
?=-+??=-+?? 解得13
43a h ?=-????=??
······················································································ 3分 ∴所求抛物线解析式为2
14
(2)3
3
y x x =--+． ····································································· 4分
（2）分三种情况：
①当02t <≤，重叠部分的面积是OPQ S △，过点A 作AF x ⊥轴于点F ，
∵(11)
A ，，在Rt OAF △中，1AF OF ==，45AOF ∠=°
在Rt OPQ △中，OP t =，45OPQ QOP ∠=∠=°，
∴cos 452
PQ OQ t ===
°， ∴2
2
11224
S t ??== ? ???． ····················································· 6分 ②当23t <≤，设PQ 交AB 于点G ，作GH x ⊥轴于点H 45OPQ QOP ∠=∠=°，则四边形OAGP 是等腰梯形，
重叠部分的面积是OAGP S 梯形． ∴2AG FH t ==-， ∴11
()(2)1122
S AG OP AF t t t =
+=+-?=-． ············ 8分 ③当34t <<，设PQ 与AB 交于点M ，交BC 于点N ，重叠部分的面积是OAMNC S 五边形．
因为PNC △和BMN △都是等腰直角三角形，所以重叠部分的面积是
OAMNC S 五边形BMN OABC S S =-△梯形．
∵(31)B ，，OP t =， ∴3PC CN t ==-，
∴1(3)4BM BN t t ==--=-，
∴2
11(23)1(4)22S t =+?--
2111
422
S t t =-+-． ······················································· 10分
（3）存在 11t = ·················································································································· 12分 22t = ················································································································ 14分
【089】解：（1）
圆心O 在坐标原点，圆O 的半径为1，
∴点A B C D 、、、的坐标分别为(10)(01)(10)(01)A B C D --，、，、，、，
抛物线与直线y x =交于点M N 、，且MA NC 、分别与圆O 相切于点A 和点C ，
∴(11)(11)M N --，、，． ·
··········································································································· 2分 点D M N 、、在抛物线上，将(01)
(11)(11)D M N --，、，、，的坐标代入 2y ax bx c =++，得：111c a b c a b c =??-=-+??=++? 解之，得：1
11a b c =-??
=??=?
∴抛物线的解析式为：21y x x =-++． ················································································ 4分
（2）
2
2
15124y x x x ?
?=-++=--+ ??
?
∴抛物线的对称轴为12
x =
，
12OE DE ∴===，． ······················· 6分
连结90BF BFD ∠=，°，
BFD EOD ∴△∽△，DE OD
DB FD
∴
=，
又122
DE OD DB ===，，
FD ∴=
，
EF FD DE ∴=-=
=． ··············································································· 8分 （3）点P 在抛物线上． ············································································································· 9分 设过D C 、点的直线为：y kx b =+，
将点(10)(01)C D ，、，的坐标代入y kx b =+，得：11k b =-=，，
∴直线DC 为：1y x =-+． ·································································································· 10分
过点B 作圆O 的切线BP 与x 轴平行，P 点的纵坐标为1y =-， 将1y =-代入1y x =-+，得：2x =．
∴P 点的坐标为(21)-，， ·
······································································································· 11分 当2x =时，2
2
12211y x x =-++=-++=-，
所以，P 点在抛物线2
1y x x =-++上． ·············································································· 12分 说明：解答题各小题中只给出了1种解法，其它解法只要步骤合理、解答正确均应得到相应的分数．
【090】（1）解：把A （1-，0），C （3，2-）代入抛物线 2
3y ax ax b =-+ 得
?
??-=+-=+-?--2990)1(3)1(2b a a b a a ···························································································· 1分
整理得 ???-==+204b b a ·················· ……………… 2分 解得???
??-==2
21b a ………………3分
∴抛物线的解析式为 22
3
212--=x x y ············································································ 4分
（2）令022
3
212=--x x 解得 1214x x =-=，
∴ B 点坐标为（4，0）
又∵D 点坐标为（0，2-） ∴AB ∥CD ∴四边形ABCD 是梯形． ∴S 梯形ABCD ＝
82)35(2
1
=?+ ·
······························· 5分 设直线)0(1≠+=k kx y 与x 轴的交点为H ，
与CD 的交点为T ，
则H （k 1
-，0）， T （k 3-，2-） ····················· 6分
∵直线)0(1≠+=k kx y 将四边形ABCD 面积二等分
∴S 梯形AHTD ＝2
1
S 梯形ABCD ＝４ ∴42)311(21=?-+-k
k ·········································· 7分 ∴34
-=k ···································································· 8分 （3）∵MG ⊥x 轴于点G ，线段MG ︰AG ＝1︰2
∴设M （m ，21+-m ）， (9)
∵点M 在抛物线上 ∴22
321212
--=+-m m m 解得1231m m ==-，（舍去） ······························· 10分
∴M 点坐标为（3，2-） ································································································ 11分 根据中心对称图形性质知，MQ ∥AF ，MQ ＝AF ，NQ ＝EF ，
∴N 点坐标为（1，3-） ······························································································· 12分
图(9) -2 图(9) -1