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 . $\qquad$
$$
-1 \leqslant a \leqslant \frac{17}{8}
$$
5. 【Analysis and Solution】Solution 1 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 $4 \cos ^{2} \theta=2(a+\sin \theta)$.
$$
\begin{aligned}
\therefore a & =2 \cos ^{2} \theta-\sin \theta=2-2 \sin ^{2} \theta-\sin \theta \\
& =-2\left(\sin \theta+\frac{1}{4}\right)^{2}+\frac{17}{8} . \\
\because-1 & \leqslant \sin \theta \leqslant 1 . \\
\therefore 0 & \leqslant\left(\sin \theta+\frac{1}{4}\right)^{2} \leqslant \frac{25}{16} .
\end{aligned}
$$

Thus $-1 \leqslant a \leqslant \frac{17}{8}$.
Solution 2 The condition of the problem is equivalent to the equation $2 y+4(y-a)^{2}=4$ having non-negative solutions.
This is the equation $y^{2}-\left(2 a-\frac{1}{2}\right) y+a^{2}-1=0$ having non-negative solutions. Therefore, we have
$$
\begin{array}{l}
\Delta=\left(2 a-\frac{1}{2}\right)^{2}-4\left(a^{2}-1\right)=\frac{17}{4}-2 a \geqslant 0, \\
2 a-\frac{1}{2}+\sqrt{\frac{17}{4}-2 a} \geqslant 0,
\end{array}
$$

Solving gives $-1 \leqslant a \leqslant \frac{17}{8}$.