已知抛物线y^2=8x,过点（a,0)且斜率为1的直线与抛物线交于不同两点A,B,且AB绝对值小于等于8,求a取值范围

设过A,B的直线方程: y = x + b
过点(a, 0): 0 = a + b, b = -a
y = x - a (1)
抛物线y^2=8x (2)
解(1)(2): x^2 -2(a + 4)x + a^2 = 0
x1 = (a + 4) - 2sqrt[2(a+2)] (sqrt:平方根)
x2 = (a + 4) + 2sqrt[2(a+2)]
A: ( (a + 4) - 2sqrt[2(a+2)], 4 - 2sqrt[2(a+2)] )
B: ( (a + 4) + 2sqrt[2(a+2)], 4 + 2sqrt[2(a+2)] )
|AB|^2 = {4sqrt[2(a+2)]}^2 + {4sqrt[2(a+2)]}^2 = 64(a+2)
AB绝对值小于等于8: |AB|^2 ≤ 64
64(a+2) ≤ 64
a ≤ -1
由判别式可知a+2 ≥ 0; a ≥ -2
a取值范围: -2 ≤ a ≤ -1
过A,B的直线斜率为1, AB的垂直平分线斜率为-1, 过AB的中M点:
(a+4, 4)
垂直平分线方程: y = -x +a + 8
点N: (a + 8, 0)
|MN| = 4sqrt(2) (三角形NAB的高)
三角形NAB的底|AB| = 8sqrt(a+2)
三角形NAB面积 = (1/2)*8sqrt(a+2)*4sqrt(2) = (16sqrt(2))*sqrt(a+2)
NAB面积的最大值为无穷大