发布时间 : 星期五 文章（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更新完毕开始阅读
∴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