(Ⅰ)由已知得:,解得a=2,b=1,c=.∴椭圆C的方程为+y2=1 (Ⅱ)由(Ⅰ)可知:A1(-2,0),A2(2,0),B1(0,1). ∴kA2B1=-. ∵l∥A1B1,∴kl=kA2B1=-. 可设直线l的方程为y=-x+m,设P(x1,y1),Q(x2,y2). 联立消去y得x2-2mx+2m2-2=0. ∵直线l与椭圆有不同的两个交点, ∴△=4m2-4(2m2-2)>0,即-<m<, ∴x1+x2=2m,x1x2=2m2-2. ∵P,Q异于椭圆C的顶点,∴α≠,β≠,∴tanα=kA1P=,tanβ=kB1Q=. ∴tanα+tanβ=+=y1x2+x1y2+2y2-x1-2 | x2(x1+2) | . ∵y1=-x1+m,y2=-x2+m. ∴tanα+tanβ=(m-1)(x1+x2)-x1x2+2m-2 | (x1+2)x2 | =2m(m-1)-(2m2-2)+2m-2 | (x1+2)x2 | =0, ∴tan(α+β)==0. 又∵α,β∈(0,π),∴α+β∈(0,2π),故α+β=π. |