利用点差法 设A(x1,y1).B(x2,y2)则A,B两点坐标满足椭圆方程x1^2/4 +y1^2/3 =1 (1)式x2^2/4 +y2^2/3 =1 (2) 式(1)减(2)式,(x1^2-x2^2)/4 +(y1^2-y2^2)/3 =0所以 (x1+x2)(x1-x2)/4 +(y1+y2)(y1-y2)/3 =0(1,1)为中点,所...