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P2H?OP2?sin?AOP2?32?12?34，?P2?，?3?4?3? ······································· 11分 ??4??3?符合条件的P点坐标有?1，?3???33，，????44???·······················································12分 ???
87.（08青海省卷28题）王亮同学善于改进学习方法，他发现对解题过程进行回顾反思，效果会更好．某一天他利用30分钟时间进行自主学习．假设他用于解题的时间x（单位：分钟）与学习收益量y的关系如图甲所示，用于回顾反思的时间x（单位：分钟）与学习收益量y的关系如图乙所示（其中OA是抛物线的一部分，A为抛物线的顶点），且用于回顾反思的时间不超过用于解题的时间． （1）求王亮解题的学习收益量y与用于解题的时间x之间的函数关系式，并写出自变量x的取值范围； （2）求王亮回顾反思的学习收益量y与用于回顾反思的时间x之间的函数关系式； （3）王亮如何分配解题和回顾反思的时间，才能使这30分钟的学习收益总量最大？ （学习收益总量?解题的学习收益量?回顾反思的学习收益量）
O 2 图甲
第28题图
（08青海省卷28题解析）解：（1）设y?kx， 把(2，4)代入，得k?2．
···································································································· （1分） ?y?2x． ·
自变量x的取值范围是：0≤x≤30．···························································· （2分） （2）当0≤x≤5时，
设y?a(x?5)?25，···················································································· （3分） 把(0，0)代入，得25a?25?0，a??1．
?y??(x?5)?25??x?10x． ································································ （5分）
222y 4 y 25 A x O 5 图乙
15 x 当5≤x≤15时，
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··········································································································· （6分） y?25 ·
??x2?10x(0≤x≤5)即y??．
?25(5≤x≤15)（3）设王亮用于回顾反思的时间为x(0≤x≤15)分钟，学习效益总量为Z， 则他用于解题的时间为(30?x)分钟． 当0≤x≤5时，
······················· （7分） Z??x?10x?2(30?x)??x?8x?60??(x?4)?76． ·
··········································································· （8分） ?当x?4时，Z最大?76． ·当5≤x≤15时，
···································································· （9分） Z?25?2(30?x)??2x?85． ·
?Z随x的增大而减小，
222?当x?5时，Z最大?75．
综合所述，当x?4时，Z最大?76，此时30?x?26． ································（10分） 即王亮用于解题的时间为26分钟，用于回顾反思的时间为4分钟时，学习收益总量最大．
······················································································································· （11分）
88.（08山东济宁26题）（12分）
△ABC中，?C?90?，?A?60?，AC?2cm．长为1cm的线段MN在△ABC的边AB上沿AB方向以1cm/s的速度向点B运动（运动前点M与点A重合）．过M，N分别作AB的垂线交直角边于
P，Q两点，线段MN运动的时间为ts．
（1）若△AMP的面积为y，写出y与t的函数关系式（写出自变量t的取值范围）；
（2）线段MN运动过程中，四边形MNQP有可能成为矩形吗？若有可能，求出此时t的值；若不可能，说明理由；
（3）t为何值时，以C，P，Q为顶点的三角形与△ABC相似？
?（08山东济宁26题解析）解：（1）当点P在AC上时，?AM?t，?PM?AM?tg60?3t．
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?y?12t?3t?322······································································ 2分 t(0≤t≤1)． ·
当点P在BC上时，PM?BM?tan30??33(4?t)．
y?12t?33(4?t)??36t?2233················································· 4分 t(1≤t≤3)． ·
（2）?AC?2，?AB?4．?BN?AB?AM?MN?4?t?1?3?t．
?QN?BN?tan30??33······································································· 6分 (3?t)． ·
由条件知，若四边形MNQP为矩形，需PM?QN，即3t??t?3433(3?t)，
．
34?当t?s时，四边形MNQP为矩形． ································································ 8分
34（3）由（2）知，当t?s时，四边形MNQP为矩形，此时PQ∥AB，
·························································································· 9分 ?△PQC∽△ABC． ·
CQCP33除此之外，当?CPQ??B?30时，△QPC∽△ABC，此时
AMAP???tan30??．
??cos60?12，?AP?2AM?2t．?CP?2?2t． ·····························10分
BN32233?BNBQ?cos30??32，?BQ??(3?t)．
又?BC?23，?CQ?23?233(3?t)?23t3． ········································ 11分
23t?13?3，t?．
2?2t3212?当t?s或
34s时，以C，P，Q为顶点的三角形与△ABC相似．···················12分
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89.（08四川巴中30题）（12分）30．已知：如图14，抛物线y??y??3434x?3与x轴交于点A，点B，与直线
34x?b与y2点C，直线y??x?b相交于点B，
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？
（08四川巴中30题解析）解：（1）在y????34x?3?0
234x?3中，令y?0 2y C E N ?x1?2，x2??2
··············································1分 ?A(?2，0)，B(2，0) ·又?点B在y???0??b?3232?b
34x?b上
A M D O P B x
34x?32?BC的解析式为y?? ············································································· 2分
32?y??x?3?x1??1??x2?2??4（2）由?，得? ···················································· 4分 9 ?y?033y??2?1?y??x??4??429??C??1，4??0) ?，B(2，?9494?AB?4，CD? ······························································································ 5分
?92?S△ABC?12?4? ························································································ 6分
（3）过点N作NP?MB于点P
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?EO?MB ?NP∥EO
?△BNP∽△BEO ······························································································· 7分 ?BNBE?NPEO·········································································································· 8分
3432由直线y??x?可得：E?0，?
?2?32?3??在△BEO中，BO?2，EO??2t52?NP3，则BE?52
，?NP?65······················································································ 9分 t·
216?S??t?(4?t)
253212S??t?t(0?t?4) ····················································································10分
553122S??(t?2)? ····························································································· 11分
5512 ?此抛物线开口向下，?当t?2时，S最大?512．····························12分 ?当点M运动2秒时，△MNB的面积达到最大，最大为5
90.（08四川自贡26题）抛物线y?ax?bx?c(a?0)的顶点为M，与x轴的交点为A、B（点B在点A的右侧），△ABM的三个内角∠M、∠A、∠B所对的边分别为m、a、b。若关于x的一元二次方程
(m?a)x?2bx?(m?a)?0有两个相等的实数根。
22（1）判断△ABM的形状，并说明理由。
（2）当顶点M的坐标为（－2，－1）时，求抛物线的解析式，并画出该抛物线的大致图形。
（3）若平行于x轴的直线与抛物线交于C、D两点，以CD为直径的圆恰好与x轴相切，求该圆的圆心坐标。
（08四川自贡26题解析）解：（1）令??(2b)?4(m?a)(m?a)?0 得a?b?m
由勾股定理的逆定理和抛物线的对称性知
△ABM是一个以a、b为直角边的等腰直角三角形
（2）设y?a(x?2)?1
∵△ABM是等腰直角三角形
∴斜边上的中线等于斜边的一半
2222215