11. 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 $\qquad$ .
$$
\left\{\begin{array}{l}
x^{2}+4(y-a)^{2}=4, \\
x^{2}=2 y
\end{array} \Rightarrow 2 y^{2}-(4 a-1) y+2 a^{2}-2=0\right.
$$

Then $\Delta=(4 a-1)^{2}-16\left(a^{2}-1\right)=17-8 a \geqslant 0$
$\Rightarrow a \leqslant \frac{17}{8}$. Therefore, the range of real number $a$ is $\left[-1, \frac{17}{8}\right]$.