点差法
设两交点(x1,y1)(x2,y2)
满足方程x1²/16+y1²/4=1 x2²/16+y2²/4=1
两式相减（x1²-x2^2)/16+(y1²-y2^2)/4=0
(y1-y2)/(x1-x2)=-(x1+x2)/4(y1+y2)=-1/2
即直线斜率为-1/2
方程为y-1=-1/2(x-2) =>x+2y-4=0