有题知c=5√2,a^2=b^2+50,即椭圆方程为y^2／a^2+x^2／(a^2-50)=1,设直线与椭圆的两个交点为m(x1,y1),n(x2,y2),带入椭圆方程得（y1）^2／a^2+(x1)^2／(a^2-50)^2,(y2)^2／a^2+(x2)^2／(a^2-50)=1,两个方程相减得(y1+y2)／(x1+x2)=-[(x2-x1)(a^2)]／[(y2-y1)(a^2-50)].①.有题知直线斜率k=(y2-y1)／(x2-x1)=3,x1+x2=2×1／2=1,y1+y2=3(x1+x2)-4=-1,代入①解得a^2=75,从而b^=25,即所求椭圆方程为x^2／25+y^2／75=1.