Example 10 (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 $\qquad$ .
Solution: Fill in $\left[-1, \frac{17}{8}\right]$. Reason: 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 $4 \cos ^{2} \theta=2(a+\sin ) \theta$, thus $a=2 \cos ^{2} \theta-\sin \theta=2-2 \sin ^{2} \theta-\sin \theta=-2\left(\sin \theta+\frac{1}{4}\right)^{2}+\frac{17}{8}$. Since $-1 \leqslant \sin \theta \leqslant 1$, we have $0 \leqslant\left(\sin \theta+\frac{1}{4}\right)^{2} \leqslant \frac{25}{16}$, hence $-1 \leqslant a \leqslant \frac{17}{8}$.