设A(a,b),B(c,d)
a^2 - 4b^2 = 4 (1)
c^2 - 4d^2 = 4 (2)
M(8,1)为线段AB的中点,8 = (a + c)/2,1 = (b+d)/2
a+c = 16,b+d = 2
(1)-(2):(a+c)(a-c) = 4(b+d)(b-d)
16(a-c) = 4*2(b-d)
(b-d)/(a-c) = 2
即AB的斜率为2,方程为 y - 1 = 2(x - 8)
y = 2x - 15