Example 20 (1998 National High School League Question) If the ellipse $x^{2}+4(y-a)^{2}=4$ intersects the parabola $x^{2}=2 y$, then the range of real number $a$ is $\qquad$ .
Solve: Fill in $-1 \leqslant a \leqslant \frac{17}{8}$. Reason: The parametric equation of the ellipse is $\left\{\begin{array}{l}x=2 \cos \theta \\ y=a+\sin \theta\end{array}\right.$ ( $\theta$ is the parameter), substituting into $x^{2}=2 y$ yields $4 \cos ^{2} \theta=2(a+\sin \theta)$.
Thus, $a=2 \cos ^{2} \theta-\sin ^{2} \theta=2-2 \sin ^{2} \theta-\sin \theta=-2\left(\sin \theta+\frac{1}{4}\right)^{2}+\frac{17}{8}$.
Since $-1 \leqslant \sin \theta \leqslant 1$, then $-1 \leqslant a \leqslant \frac{17}{8}$.