把y=mx+1代入x^2/4+y^2/b=1化简得(4m^2+b)x^2+8mx+4-4b=0
由直线l与椭圆C恒有公共点则△=64m^2-4(4m^2)(4-4b)≥0
化简得b(b+4m^2-1)≥0 b1=0 ,b2=1-4m^2
当1-4m^2>0即-1/2
把y=mx+1代入x^2/4+y^2/b=1化简得(4m^2+b)x^2+8mx+4-4b=0
由直线l与椭圆C恒有公共点则△=64m^2-4(4m^2)(4-4b)≥0
化简得b(b+4m^2-1)≥0 b1=0 ,b2=1-4m^2
当1-4m^2>0即-1/2