（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．
2?y??(x?5)2?25??x2?10x．························································ （5分）
当5≤x≤15时，
···························································································· （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，
Z??x2?10x?2(30?x)??x2?8x?60??(x?4)2?76． ····················· （7分） ?当x?4时，Z最大?76． ································································· （8分）
当5≤x≤15时，
Z?25?2(30?x)??2x?85． ··························································· （9分）
Z随x的增大而减小，
?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；
t为何值时，（3）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?AMtg60?3t．
1?y?t23t?32t(0≤t≤1)． ····························································· 2分 23(4?t)． 3当点P在BC上时，PM?BMtan30?133223y?t(4?t)??t?t(1≤t≤3)． ··········································· 4分
2363（2）
AC?2，?AB?4．?BN?AB?AM?MN?4?t?1?3?t．
3(3?t)． ······························································ 6分 33(3?t)， 3?QN?BNtan30?由条件知，若四边形MNQP为矩形，需PM?QN，即3t??t?3． 43?当t?s时，四边形MNQP为矩形． ······················································· 8分
43（3）由（2）知，当t?s时，四边形MNQP为矩形，此时PQ∥AB，
4?△PQC∽△ABC． ·············································································· 9分
除此之外，当?CPQ??B?30时，△QPC∽△ABC，此时
CQ3?tan30?． CP3AM1?cos60?，?AP?2AM?2t．?CP?2?2t． ························· 10分 AP2BN23BN3，?BQ??(3?t)． ?cos30?3BQ232又
BC?23，?CQ?23?2323t(3?t)?． ··································· 11分 3323t31?3?，t?． 2?2t32?当t?13s或s时，以C，P，Q为顶点的三角形与△ABC相似． ················ 12分 24
89.（08四川巴中30题）（12分）30．已知：如图14，抛
32x?3与x轴交于点A，点B，与直线433点C，直线y??x?b与yy??x?b相交于点B，
44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
设运动BC上以每秒2个单位长度的速度从B向C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
物线y??大面积是多少？
（08四川巴中30题解析）解：（1）在y??32x?3中，令y?0 4C E y 3??x2?3?0
4?x1?2，x2??2
N ?A(?2，0)，B(2，0) ········································ 1分
又
点B在y??A M D O P B x 3x?b上 43?0???b
23b?
233·································································· 2分 ?BC的解析式为y??x? ·
4232?y??x?3?x1??1???4（2）由?，得?9
y1??y??3x?3??4??429???C??1，?，B(2，0)
4???x2?2 ············································· 4分 ??y2?09················································································ 5分 ?AB?4，CD? ·4199··········································································· 6分 ?S△ABC??4?? ·
242（3）过点N作NP?MB于点P EO?MB ?NP∥EO
················································································· 7分 ?△BNP∽△BEO ·
BNNP ··························································································· 8分 ??BEEO由直线y??33?3?x?可得：E?0，? 42?2?35，则BE? 22?在△BEO中，BO?2，EO?62tNP，?NP?t ·········································································· 9分 ?5352216?S?t(4?t)
25312······································································· 10分 S??t2?t(0?t?4) ·
55312··············································································· 11分 S??(t?2)2? ·
5512此抛物线开口向下，?当t?2时，S最大?
512?当点M运动2秒时，△MNB的面积达到最大，最大为． ························ 12分
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
2(m?a)x2?2bx?(m?a)?0有两个相等的实数根。
（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 又顶点M(－2，－1) ∴
222221AB?1，即AB＝2 22∴A(－3，0)，B(－1，0)
将B(－1，0) 代入y?a(x?2)?1中得a?1
∴抛物线的解析式为y?(x?2)?1，即y?x?4x?3 图略
（3）设平行于x轴的直线为y?k
解方程组错误！不能通过编辑域代码创建对象。 得x1??2?22k?1，x2??2?k?1 （k??1)
∴线段CD的长为2k?1 ∵以CD为直径的圆与x轴相切 据题意得k?1?k ∴k?k?1 解得 k?21?5 21?51?5)和(?2,) 22∴圆心坐标为(?2,
91.（08新疆自治区24题）（10分）某工厂要赶制一批抗震救灾用的大型活动板房．如图，板房一面的形状是由矩形和抛物线的一部分组成，矩形长为12m，抛物线拱高为5.6m． （1）在如图所示的平面直角坐标系中，求抛物线的表达式．
（2）现需在抛物线AOB的区域内安装几扇窗户，窗户的底边在AB上，每扇窗户宽1.5m，高1.6m，相邻窗户之间的间距均为0.8m，左右两边窗户的窗角所在的点到抛物线的水平距离至少为0.8m．请计算最多可安装几扇这样的窗户？
2024-2024年中考数学压轴题精选(九)及答案资料
