10. Given the ellipse $\frac{y}{4}+x^{2}=1, P$ is any point on the ellipse. Through point $P$, draw lines parallel to $l_{1}: y=2 x$ and $l_{2}: y=-2 x$ which intersect lines $l_{2}, l_{1}$ at points $M, N$ respectively. Then the maximum value of $|M N|$ is $\qquad$
10.2.

As shown in Figure 5.
Let $P\left(x_{P}, y_{P}\right)$.
Thus, the lines parallel to $l_{1}$ and $l_{2}$ are $l_{3}: y=2\left(x-x_{P}\right)+y_{P}$, $l_{4}: y=-2\left(x-x_{P}\right)+y_{P}$.
From $2 x-2 x_{P}+y_{P}=-2 x$ $\Rightarrow x_{M}=\frac{2 x_{P}-y_{P}}{4}, y_{M}=-\frac{2 x_{P}-y_{P}}{2}$.
From $-2 x+2 x_{P}+y_{P}=2 x$
$\Rightarrow x_{N}=\frac{2 x_{P}+y_{P}}{4}, y_{N}=\frac{2 x_{P}+y_{P}}{2}$.
Therefore, $M N^{2}=\left(\frac{1}{4} \times 2 y_{P}\right)^{2}+\left(\frac{1}{2} \times 4 x_{P}\right)^{2}$ $=\frac{1}{4} y_{P}^{2}+4 x_{P}^{2}$.
Also, $\frac{1}{4} y_{P}^{2}+x_{P}^{2}=1$, then
$$
M N^{2}=1+3 x_{P}^{2} \leqslant 4 \text {. }
$$

Thus, $M N \leqslant 2$, when $P(1,0)$ or $(-1,0)$, $M N=2$.