解:F1(-1,0),F2(-1,0)c=1设A(x1,y1),B(x2,y2),直线AB方程是 x=my+1△F1AB的面积S=(1/2)*2c*|y1-y2|=|y1-y2|由 x=my+1和x^2/3+y^2/2=1消去x并化简得:(2m^2+3)y^2+4my-4=0△=(4m)^2-(2m^2+3)(-4)=12(2m^2+1)S=|y1-y2...