(1)x^2/2+y^2=1(x≠±根号2,y≠0)
（2）设l的方程为：x=ty+1与x^2/2+y^2=1联立
消去x得：（ty+1)^2+2y^2-2=0
即 （t^2+2)y^2+2ty-1=0
设M(x1,y1),N(x2,y2),MN中点为Q(x',y')
韦达定理：
y1+y2=-2t/(t^2+2),y1y2=-1/(t^2+2)
y'=-t/(t^2+2),x'=ty'+1=2/(t^2+2)
MN的垂直平分线m的方程为：
y+t/(t^2+2)=-t[x-2/(t^2+2)]
令x=0,得 y=t/(t^2+2)
∴m与y轴交点T(0,t/(t^2+2)
∵以MN为对角线的正方形的第三个顶点恰在Y轴上
∴T即是该点∴TM⊥TN
∴向量TM·向量TN=0
(x1,y1-t/(t^2+2) )·(x2,y2-t/(t^2+2))=0
x1x2+[y1-t/(t^2+2)][y2-t/(t^2+2)]=0
(ty1+1)(ty2+1)+y1y2-t/(t^2+2)×(y1+y2)+t^2/(t^2+2)^2=0
(t^2+1)y1y2+[t-t/(t^2+2)](y1+y2)+t^2/(t^2+2)^2+1=0
-(t^2+1)/(t^2+2)+(t^3+t)/(t^2+2)×(-2t)/(t^2+2)+t^2/(t^2+2)^2+1=0
==>t^4=1==>t=±1
∴ 直线L方程为x= ±y+1 即x+y-1=0或x-y-1=0