Calculus (3rd Edition)

The intersection is empty if and only if $h\in[-\frac{1}{2},\frac{1}{2}]$.
Putting $z=h$ in the equation, we have $$-x^2-4y^2+4h^2=1$$ hence we get $$x^2+4y^2=1-4h^2.$$ The intersection is non-empty whenever $$4h^2-1\gt 0\Longrightarrow(h+\frac{1}{2})(h-\frac{1}{2})\gt 0$$ hence $$h\in (-\infty,-\frac{1}{2})\cup (\frac{1}{2},\infty) .$$ So, the intersection is empty if and only if $h\in[-\frac{1}{2},\frac{1}{2}]$.