设A、B关于直线l:y=4x+m对称,∴直线l就是线段AB的中垂线.∴AB的斜率k*4=-1,k=-1/4
设A(x1,y1),B(x2,y2),中点M(x0,y0)
则x1+x2=2x0,y1+y2=2y0,
且k=(y1-y2)/(x1-x2)=-1/4,即x1-x2= - 4(y1-y2)
把A(x1,y1),B(x2,y2)代入椭圆方程得
x1²/4 + y1²/3=1.①
x2²/4 + y2²/3=1.②
①-②得
(x1-x2)²/4 + (y1-y2)²/3=0
(x1-x2)(x1+x2)/4 + (y1-y2)(y1+y2)/3=0
把x1-x2= - 4(y1-y2)代入得
-(x1+x2) + (y1+y2)/3=0
即y1+y2=3(x1+x2)
又∵x1+x2=2x0,y1+y2=2y0,代入得
y0=3x0.③
中点M(x0,y0)在直线l:y=4x+m上,代入得
y0=4x0+m.④
由③④解得
x0= - m,y0= - 3m
∵中点M肯定在椭圆内部
∴满足x0²/4 + y0²/3＜1
把x0= - m,y0= - 3m代入得
m²/4 + 9m²/3＜1
∴ - (2√13)/13＜m＜(2√13)/13