Example 2 If the ellipse $x^{2}+4(y-a)^{2}=4$ and the parabola $x^{2}=2 y$ have common points, then the range of the real number $a$ is $\qquad$
(1998, National High School Mathematics Competition)
Solution: From $x^{2}+4(y-a)^{2}=4$, we can set
$$
\begin{array}{l}
x=2 \cos \theta, y=a+\sin \theta . \\
\text { Substituting into } x^{2}=2 y, \text { we get } \\
4 \cos ^{2} \theta=2(a+\sin \theta) . \\
\text { Therefore, } a=2 \cos ^{2} \theta-\sin \theta \\
\quad=2-2 \sin ^{2} \theta-\sin \theta \\
\quad=-2\left(\sin \theta+\frac{1}{4}\right)^{2}+\frac{17}{8} . \\
\because-1 \leqslant \sin \theta \leqslant 1, \\
\therefore 0 \leqslant\left(\sin \theta+\frac{1}{4}\right)^{2} \leqslant \frac{25}{16} .
\end{array}
$$

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