13. (25 points) 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?
13. It is known that $l_{T M}: y=\frac{x}{t}+1$.

Substituting into the ellipse equation, we get $E\left(\frac{-8 t}{t^{2}+4}, \frac{t^{2}-4}{t^{2}+4}\right)$. Similarly,
$$
l_{T N}: y=\frac{3 x}{t}-1, F\left(\frac{24 t}{t^{2}+36}, \frac{-t^{2}+36}{t^{2}+36}\right) .
$$

Then $l_{E F}: y-\frac{t^{2}-4}{t^{2}+4}=-\frac{t^{2}-12}{16 t}\left(x+\frac{8 t}{t^{2}+4}\right)$, which simplifies to $\frac{t^{2}-12}{16 t} x+y-\frac{1}{2}=0$.
Thus, the line $E F$ passes through the fixed point $\left(0, \frac{1}{2}\right)$.
Therefore, $S_{\triangle T E F}=S_{\triangle T M N}-S_{\triangle F M N}+S_{\triangle M E F}$
$=\frac{1}{2} \times 2|t|-\frac{1}{2} \times 2\left|\frac{24 t}{t^{2}+36}\right|+$
$\frac{1}{2} \times \frac{1}{2}\left|\frac{24 t}{t^{2}+36}+\frac{8 t}{t^{2}+4}\right|$
$=|t|-\left|\frac{24 t}{t^{2}+36}\right|+|2 t|\left|\frac{3}{t^{2}+36}+\frac{1}{t^{2}+4}\right|$.
Also, $S_{\triangle T M N}=\frac{1}{2} \times 2|t|=|t|$, so
$k=\frac{S_{\triangle T M N}}{S_{\triangle T E F}}=\frac{t^{4}+40 t^{2}+144}{t^{4}+24 t^{2}+144}$
$=1+\frac{16}{t^{2}+\frac{144}{t^{2}}+24}$
$$
\leqslant 1+\frac{16}{2 \sqrt{144}+24}=\frac{4}{3} \text {. }
$$

When and only when $t^{2}=\frac{144}{t^{2}}$, i.e., $t= \pm 2 \sqrt{3}$, $k$ reaches its maximum value $\frac{4}{3}$.