问一道初三关于抛物线、相似形的压轴题（有追加）（急）

(1) 代入两点的坐标：
A(-2, 4): 4m - 4m + n = 4, n = 4
B(1, 0): m + 2m + n = 0, 3m = -n = -4, m = -4/3
(2)原抛物线方程为 y = -4x²/3 -8x/3 + 4
= -4(x+1)²/3 + 16/3
AB = √[(-2-1)² + (4-0)²] = 5
原抛物线向右平移,则AA'则与x轴(及BB')平行, 所以只需AA'=AB即可.
于是A'(3, 4) (纵坐标不变,横坐标差为5, 即AA')
y= -4(x+1)²/3 + 16/3向右平移后,新的方程可由原抛物线方程中的x变为x-5得到:
y = -4(x-5+1)²/3 + 16/3
= -4(x-4)²/3 + 16/3
(3)新抛物线的对称轴为x = 4
B向右平移5个单位, B'(6, 0)
AB'的方程为: y/((x-6) = (4-0)/(-2-6) = -1/2
x + 2y = 6
x = 4, y = 1
C(4, 1)
AA'B'B为菱形,角BAC = 角A'AC = 角DB'C
所以要使而二者相似,只需再有两个对应角相等即可.
在∆ABC中, AC的斜率kAC = ((4-1)/(-2-4) = -1/2
BC的斜率kBC = ((0-1)/(1-4) = 1/3
tanABC = |(kAC - kBC)/(1 + kAC*kBC)|
= |(-1/2 -1/3)/(1 -(1/2)(1/3)| = 1
角ACB = 45°
下面 有两种情况：角CDB' = 45°或角DCB' = 45°
(i) 角CDB' = 45°
CD的斜率=tan45° = 1
设CD方程: y = x + b
代入C(4, 1), b = -3
y = x - 3
取y = 0, x = 3
D(3, 0)
(ii) 角DCB' = 45°
从图中容易看出,CD的斜率为负, 设为k
AC即CB'的斜率kAC = -1/2
tan角DCB' = tan45° = 1 = |(kAC - k)/(1+k*kAC)|
|(-1/2 -k)/(1 -k/2)| = 1
|(1 + 2k)/(2 - k)| = 1
(1 + 2k)/(2 - k) = ±1
k = 1/3 (>0, 舍去)
k = -3
设CD方程: y = -3x + b
代入C(4, 1), b = 13
y = -3x + 13
取y = 0, x = 13/3
D(13/3, 0)
注：图是用计算机画的.原来用的计算机不能画,现在补上.

解得m=-4/3n=4；

（2）四边形AA′B′B为菱形，

则AA′=B′B=AB=5；

∵y=-4/3x^2-8/3x+4，

=-4/3(x+1)2+16/3；

∴向右平移...