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
11. $\left[-1, \frac{17}{8}\right]$ Hint: Substitute $x^{2}=2 y$ into $x^{2}+4(y-a)^{2}=4$, the problem is equivalent to: The equation $4(y-a)^{2}+2 y$ $=4$ has non-negative real solutions $\Leftrightarrow y=\frac{(4 a-1)+\sqrt{17-8 a}}{4} \geqslant 0$, solving this, we get $a \in\left[-1, \frac{17}{8}\right]$.