由题意，将A(31)，，D
31
，代入y22
ax
2
bx2中得
3a34a
3b32b
22
112
解得
ab
89
89539
x
2
所求抛物线表达式为：
y
539
x2················································9分
（3）存在符合条件的点理由如下：Q矩形
····························································· 10分P，点Q． ·
ABOC的面积
ABgBO323．
以O，B，P，Q为顶点的平行四边形面积为由题意可知又QOB
OB为此平行四边形一边，
3
OB边上的高为2 ···················································································· 11分
依题意设点
2)P的坐标为(m，
89
539
Q点P在抛物线y
x
2
x2上
89
m
2
539
m22
解得，m10，m2
538
538，2
P，2)，P21(0
Q以O，B，P，Q为顶点的四边形是平行四边形，PQ∥OB，PQ
OB
3，
y E
A B F
C
D
x
当点P1的坐标为(0，2)时，点Q的坐标分别为
Q1(3，2)，Q2(3，2)；
O M
当点P2的坐标为
538
，2时，
点Q的坐标分别为Q3
1338
，2，Q4
338
，······································· 14分2． ·
4、（08辽宁12市26题）（本题14分）26．如图16，在平面直角坐标系中，直线y
3x3
与x轴交于点A，与y轴交于点C，抛物线yax
2
233
xc(a0)经
过
A，B，C三点．
（1）求过A，B，C三点抛物线的解析式并求出顶点F的坐标；（2）在抛物线上是否存在点P，使△ABP为直角三角形，若存在，直接写出P点坐标；若不存在，请说明理由；（3）试探究在直线AC上是否存在一点M，使得△MBF的周长最小，若存在，求出M点的坐标；若不存在，请说明理由．
y
A O B
x
C
F
图16
解：（1）Q直线y3x
3与x轴交于点A，与y轴交于点C．
A(1，0)，C(0，3)·············································································1分
Q点A，C都在抛物线上，
0
a3
233c
c
ac33
2
333233
抛物线的解析式为
yxx3·················································3分
，顶点F1
433
··················································································4分
（2）存在 ·······························································································5分
P1(0，3)······························································································7分
P3) ·····························································································9分2(2，
（3）存在 ····························································································· 10分理由：解法一：延长过点
使BCBC到点B，
BC，连接BF交直线AC于点M，则点M就是所求的点．
········································································· 11分
y
B作BHAB于点H．
33x
2
QB点在抛物线y
233
x3上，B(3，0)
H
x
在
Rt△BOC中，tanOBC
OBC
30，BC
o
33
，
B
A C
O M F
B
23，
图9
在
Rt△BBH中，BH
3BH
6，
1
BB2OH
3，
23，
B(3，23) ········································ 12分
BH
设直线
BF的解析式为y
3kk
bb
kxb
36332
234333k
解得
b
yx
33 ················································································· 13分
62y3x3
y
36
x
332
在直线
AC上存在点x
3
解得
7103M
37
，
103y
7
7，
M，使得△MBF的周长最小，此时
M
31037
，
7
·· 14分
．
2024学年中考数学压轴题精选（二次函数）（16题）附详细解答和评分标准
