c=1,设椭圆方程为x^2/(b^2+1)+y^2/b^2=1,
把y=x-√3代入上式得
b^2x^2+(b^2+1)(x^2-2√3x+3)=b^4+b^2,
(2b^2+1)x^2-2√3（b^2+1)x+3+2b^2-b^4=0,
因椭圆与直线相切,故
△/4=3（b^2+1)^2-(2b^2+1)(3+2b^2-b^4)
=.3b^4+6b^2+3
+2b^6-4b^4-6b^2
.+b^4-2b^2-3
=2b^6.-2b^2=0,b^2>0,
∴b^4=1,b^2=1.
∴椭圆方程为x^2/2+y^2=1.