5. If the ellipse $x^{2}+4(y-a)^{2}=4$ intersects the parabola $x^{2}=2 y$, then the range of the real number $a$ is $\qquad$
5. $-1 \leqslant a \leqslant \frac{17}{8}$.

From $x^{2}+4(y-a)^{2}=4$, we can set $x=2 \cos \theta, y=a+ \sin \theta$, substituting into $x^{2}=2 y$ gives $4 \cos ^{2} \theta=2(a+\sin \theta)$.
$$
\begin{aligned}
\therefore a & =2 \cos ^{2} \theta-\sin \theta=2-2 \sin ^{2} \theta-\sin \theta \\
& =-2\left(\sin \theta+\frac{1}{4}\right)^{2}+\frac{17}{8} . \\
\because & -1 \leqslant \sin \theta \leqslant 1,
\end{aligned}
$$
$$
\therefore 0 \leqslant\left(\sin \theta+\frac{1}{4}\right)^{2} \leqslant \frac{25}{16} \text {. }
$$

Thus, $-1 \leqslant a \leqslant \frac{17}{8}$.