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(1)F1(-c,0),F2(c,0)直线l的方程：y=√3(x-c)F1到直线l的距离为2√32√3=|√3(-c-c)-0|/√(3+1)=|√3c|，c=2椭圆C的焦距为4(2)设出A(x1,y1),B(x2,y2),F2(c,0),因为AF2=2F2B,即(c-x1,-y1)=2(x2-c,y2), 即y1=-2y2x^2/a^2+y^2/b^2=1与y=√3(x-c)联立，得到(1/3b^2+a^2)y^2+(2b^2c/√3)y-b^4=0y1+y2=-(2b^2c/√3)/ (1/3b^2+a^2)=-y2, y2=(2b^2c/√3)/ (1/3b^2+a^2)y1*y2=-b^4/(... |
