已知椭圆x^2+2y^2=4,
设以（1,1）为中点的弦交圆为
A(x1,y1),B(x2,y2)
则x1^2+2y1^2=4.(1)
x2^2+2y2^2=4.(2)
(1)-(2)
(x1^2-x2^2)+2[y1^2-y2^2]=0
(x1+x2)(x1-x2)=-2(y1+y2)(y1-y2)
以（1,1）为中点的弦交圆
(x1+x2)/2=1
(y1+y2)/2=1
x1+x2=2.y1+y2=2
则(y1-y2)/(x1-x2)=-2
即所求直线的斜率为-2
且过(1,1)
则y-1=-2(x-1)
2x+y-3=0