(Ⅰ)∵F(-c,0)在直线l:x-y+1=0上,∴-c+1=0,即c=1,又e= c a = 1 2 ,∴a=2c=2,∴b= a2?c2 = 22?12 = 3 .从而椭圆E的方程为 x2 4 + y2 3 =1.(Ⅱ)由e= c a = 1 2 ,得c= 1 2 a,∴b= a2?c2 = a2? a2 4 = 3 a 2 ,椭圆E的方