假设M(x1,y1),N(x2,y2),MN中点为P(x0,y0),则x1 + x2 = 2x0 ,y1 + y2 = 2y0
将M、N坐标代入椭圆方程相减可得：(-4)[(y1)^2 - (y2)^2] = 3[(x1)^2 - (x2)^2]
即：[(y1 - y2)/(x1 - x2)]·[(y1 + y2)/(x1 + x2)] = -3/4
∵M、N在y = kx + m上 ,∴k = [(y1 - y2)/(x1 - x2)] ,∴上式化为：k·(y0/x0) = -3/4
∵PG垂直平分MN ,∴K(PG) = -1/k = y0/[x0 - (1/8)]∴k = [(1/8) - x0]/y0 = (-3x0)/(4y0)
∴x0 = 1/2 ,∴k = -3/(8y0) ,y0 = -3/(8k)
又∵P(x0,y0)在y = kx + m上 ,∴y0 = kx0 + m
∴m = [-3/(8k)] - (k/2)
∴直线l为：y = kx + [-3/(8k)] - (k/2)
代入椭圆方程：3x^2 + 4{kx + [-3/(8k)] - (k/2)}^2 = 12
即：(3 + 4k^2)x^2 + 4x·[(-3/4) - k^2] + 4{[3/(8k)] + (k/2)}^2 - 12 = 0
∵M、N是不同的两点,∴△>0
∴16[(-3/4) - k^2]^2 > 4(3 + 4k^2)[4{[3/(8k)] + (k/2)}^2 - 12]
∴[(3/4) + k^2]^2 > (3 + 4k^2)[{[3/(8k)] + (k/2)}^2 - 3]
∴4(3 + 4k^2) > [3/(8k)] + (k/2)}^2 - 3 = 9/(64k^2) + (k^2/4) + (3/8) - 3
令t = k^2 > 0 ,则12 + 16t > 9/(64t) + (t/4) - (21/8)
即：12·64t + 16·64t^2 > 9 + 16t^2 - 168t
∴16·63t^2 + 12·78t - 9 > 0
∴112t^2 + 104t - 1 > 0
解得t (-13 + 4√11)/28
即 k^2 > (4√11 - 13)/28
∴k的范围是：k > {√[(28√11) - 91]}/14 ,或k < -{√[(28√11) - 91]}/14
