jmc

We try to solve the two equations $x^2+4y^2=4$ and $x^2-m(y+2{)}^2=1$ simultaneously. To eliminate $x,$ we can subtract the second equation from the first equation, giving $$4y^2+m(y+2{)}^2=3,$$or $$(m+4)y^2+(4m)y+(4m-3)=0.$$For the ellipse and hyperbola to be tangent, this equation must have exactly one solution for $y,$ so its discriminant must be zero: $$(4m{)}^2-4(m+4)(4m-3)=0,$$which simplifies to $$48-52m=0.$$Thus, $m=\boxed{\frac{12}{13}}.$