用点差法设两交点A(X1,Y1)B(X2,Y2),则x1+x2=2,y1+y2=2代入椭圆方程x1^2/4+y1^2/3=1x2^2/4+y2^2/3=1相减(x1+x2)(x1-x2)/4+(y1+y2)(y1-y2)/3=02(x1-x2)/4+2(y1-y2)/3=0得AB斜率K=(y1-y2)/(x1-x2)=-3/4所以直线方程为Y-...