(9月最新修订版)2011全国各地中考数学试题分类汇编考点18-2二次函数的应用(几何)3 下载本文

∴Q点的坐标是?8???6t168?t,t?,∴PE=8+-t=8+t,∴

5555?S?②当

114?1?216?MP?PE??t??8?t??t2?t. 223?5?1535< t ≤3时,如图2,过点Q作QF⊥x轴于F,∵BQ=2t-5,∴OF=11-(2t-5)=16-2t, 2∴Q点的坐标是(16-2t,4),∴PF=16-2-t=16-3 t .

11432?MP?PF??t??16-3t??-2t2?t. 223316③当点Q与点M相遇时,16-2 t= t,解得t=.

316当3< t <时,如图3,MQ=16-2t - t =16-3t,MP=4,

311∴S??MP?MQ??4??16-3t??-6t?32.

222216252160t?t??t?20?-(3)①当0 < t ≤时,S?.

21531532∵a=>0时,抛物线开口向上,对称轴为直线t =-20,

158555当0< t ≤时,S随t的增大而增大,∴当t=时,S有最大值.最大值为.

226∴S?8532?8?128②当< t≤3时,S?-2t2?,∵a=-2<0,抛物线开口向下,∴当t=t?-2?t-??2339?3?时,S有最大值,最大值为③当3< t <

2128. 91616时,S = -6t+32,∵k=-6<0,∴S随着t的增大而减小,又∵当t=3时,S=14,当t=338128时,S有最大值,最大值为. 39时,S=0,所以0

60时,△QMN为等腰三角形. 139. (2011湖北襄阳,26,13分)

如图10,在平面直角坐标系xOy中,AB在x轴上,AB=10,以AB为直径的⊙O′与y轴正半

轴交于点C,连接BC,AC.CD是⊙O′的切线,AD⊥CD于点D,tan∠CAD=

y?ax2?bx?c过A,B,C三点.

1,抛物线2(1)求证:∠CAD=∠CAB; (2)①求抛物线的解析式;

②判定抛物线的顶点E是否在直线CD上,并说明理由;

(3)在抛物线上是否存在一点P,使四边形PBCA是直角梯形.若存在,直接写出点P的坐标(不写求解过程);若不存在,请说明理由.

yDCAO'OBx图10 【答案】

(1)证明:连接O′C. ∵CD是⊙O′的切线,∴O′C⊥CD ··························································································· 1分 ∵AD⊥CD,∴O′C∥AD,∴∠O′CA=∠CAD ········································································· 2分 ∵O′C=O′A,∴∠O′CA=∠CAB ∴∠CAD=∠CAB.·································································································································· 3分 (2)∵AB是⊙O′的直径,∴∠ACB=90° ∵OC⊥AB,∴∠CAB=∠OCB,∴△CAO∽△BCO,∴OCOB ?OAOC即OC2?OA?OB ···························································································································· 4分 ∵tan∠CAO=tan∠CAD=

1,∴OA=2OC 2又∵AB=10,∴OC2?2OC?(10?2OC) ∵OC>0

∴OC=4,OA=8,OB=2. ∴A(-8,0),B(2,0),C(0,4) ············································································· 5分 ∵抛物线y?ax2?bx?c过A,B,C三点.∴c=4

1?a????4a?2b?4?0?4由题意得?,解之得?,

364a?8b?4?0??b???2?∴y??123················································································································· 7分 x?x?4. ·

42O'FO'C ?AFAD(3)设直线DC交x轴于点F,易证△AOC≌△ADC,∴AD=AO=8.

∵O′C∥AD,∴△FO′C∽△FAD,∴

∴8(BF+5)=5(BF+10),∴BF?1610,∴F(, ·············································· 8分 0). ·

3,33??m?4?k???设直线DC的解析式为y?kx?m,则?16,即?4

k?m?0??3?m?4?3∴y??x?4. ··························································································································· 9分

413125由y??x2?x?4??(x?3)2?得

4244顶点E的坐标为E(?3,将E(?3,25································································································ 10分 ) ·4253)代入直线DC的解析式y??x?4中, 44325右边???(?3)?4??左边.

44∴抛物线的顶点E在直线CD上. ····················································································· 11分 (3)存在.P1(?10,?6),P2(10,?36) ························································································ 13分

yDECAO'OBFx

10.(2011贵州遵义,27,14分)已知抛物线y?ax2?bx?3(a?0)经过A(3,0), B(4,1)两点,且

与y轴交于点C。 (1)求抛物线y?ax?bx?3(a?0)的函数关系式及点C的坐标; (2)如图(1),连接AB,在题(1)中的抛物线上是否存在点P,使△PAB是以AB为直角边的直

角三角形?若存在,求出点P的坐标;若不存在,请说明理由;

(3)如图(2),连接AC,E为线段AC上任意一点(不与A、C重合)经过A、E、O三点的圆交

直线AB于点F,当△OEF的面积取得最小值时,求点E的坐标。

2

【答案】(1)

A(3,0)、B(4,1)代入y=ax2+bx+3中?0=9a+3b?3??1?16a?4b?3?1a=??解得?2?b=-5??2125∴解析式为y=x-x+322令x=0时,y=3∴C点坐标为(0,3)(2)若∠PAB=90°,分别过P、B作x轴的垂线,垂足分别为E、F。

图(1) 易得△APE∽△BAF,且△BAF为等腰直角三角形,∴△APE为等腰直角三角形。 设PE=a,则P点的坐标为(a,a-3)代入解析式

3-a=

125a?a?3 解得a=0,或a=3(与A重合舍去) 22