13. Given the ellipse $C: \frac{x^{2}}{4}+y^{2}=1$ with its upper vertex $M$ and lower vertex $N$, and $T(t, 2)(t \neq 0)$ is a point on the line $y=2$. The lines $T M$ and $T N$ intersect the ellipse $C$ at points $E$ and $F$ respectively. If the area of $\triangle T M N$ is $k$ times the area of $\triangle T E F$, for what value of $t$ is $k$ maximized?
Given $E\left(x_{1}, y_{1}\right), F\left(x_{2}, y_{2}\right), E F: y=m x+n$,
then $k_{T M}=\frac{y_{1}-1}{x_{1}}=\frac{1}{t}, k_{T N}=\frac{y_{2}+1}{x_{2}}=\frac{3}{t}$, thus $\frac{y_{2}+1}{x_{2}}=\frac{3\left(y_{1}-1\right)}{x_{1}} \Rightarrow x_{1}\left(y_{2}+1\right)=3 x_{2}\left(y_{1}-1\right)$.
Also, $\frac{x_{1}^{2}}{4}+y_{1}^{2}=1 \Rightarrow x_{1}^{2}=-4\left(y_{1}-1\right)\left(y_{1}+1\right)$
$\Rightarrow 3 x_{1} x_{2}=-4\left(y_{1}+1\right)\left(y_{2}+1\right)=-4\left(m x_{1}+n+1\right)\left(m x_{2}+n+1\right)$
$\Rightarrow\left(4 m^{2}+3\right) x_{1} x_{2}+4 m(n+1)\left(x_{1}+x_{2}\right)+4(n+1)^{2}=0$.
Solving the system $\left\{\begin{array}{l}y=m x+n, \\ x^{2}+4 y^{2}=4\end{array} \Rightarrow\left(4 m^{2}+1\right) x^{2}+8 m n x+4 n^{2}-4=0\right.$
$\Rightarrow x_{1}+x_{2}=-\frac{8 m n}{4 m^{2}+1}, x_{1} x_{2}=\frac{4 n^{2}-4}{4 m^{2}+1}$, substituting into $(*)$ gives
$(n+1)\left[\left(4 m^{2}+3\right)(n-1)-8 m^{2} n+(n+1)\left(4 m^{2}+1\right)\right]=0 \Rightarrow n=\frac{1}{2}$.
Solving the system $\left\{\begin{array}{l}x=t y-t, \\ x^{2}+4 y^{2}=4\end{array} \Rightarrow\left(t^{2}+4\right) y^{2}-2 t^{2} y+t^{2}-4=0 \Rightarrow E\left(-\frac{8 t}{t^{2}+4}, \frac{t^{2}-4}{t^{2}+4}\right)\right.$,
$$
\left\{\begin{array}{l}
3 x=t y+t, \\
x^{2}+4 y^{2}=4
\end{array} \Rightarrow\left(t^{2}+36\right) y^{2}+2 t^{2} y+t^{2}-36=0 \Rightarrow F\left(\frac{24 t}{t^{2}+36}, \frac{36-t^{2}}{t^{2}+36}\right)\right. \text {. }
$$

Thus, $k_{E F}=-\frac{t^{2}-12}{16 t} \Rightarrow E F: y=-\frac{t^{2}-12}{16 t} x+\frac{1}{2}$, setting $y=2 \Rightarrow x=-\frac{24 t}{t^{2}-12}$.
Therefore, $S_{\triangle T E F}=\frac{1}{2}\left|\left(-\frac{24 t}{t^{2}-12}-t\right)\left(\frac{36-t^{2}}{t^{2}+36}-\frac{t^{2}-4}{t^{2}+4}\right)\right|=\frac{|t|\left(t^{2}+12\right)^{2}}{\left(t^{2}+4\right)\left(t^{2}+36\right)}$.
And $S_{\triangle T M N}=|t|$, thus $k=\frac{S_{\triangle T M N}}{S_{\triangle T E F}}=\frac{\left(t^{2}+4\right)\left(t^{2}+36\right)}{\left(t^{2}+12\right)^{2}} \xlongequal{u=t^{2}+12} \frac{(u-8)(u+24)}{u^{2}}$ $=-\frac{8 \times 24}{u^{2}}+\frac{16}{u}+1 \leqslant \frac{4}{3}$, so when $u=24 \Rightarrow t= \pm 2 \sqrt{3}$, $k$ reaches its maximum value.