14. $l$ is a tangent line of the hyperbola $a: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \quad(a, b>0)$, with the point of tangency being $T$. Line $l$ intersects the asymptotes of $a$ at points $A$ and $B$, and the midpoint of segment $AB$ is $M(2,1)$.

The ellipse $\frac{x^{2}}{a_{0}^{2}}+\frac{y^{2}}{b_{0}^{2}}=1 \quad\left(a_{0}, b_{0}>0\right)$ has two endpoints $C\left(0, b_{0}\right)$ and $D\left(a_{0}, 0\right)$, and $O$ and $T$ are on opposite sides of $CD$. The angle bisectors (internal and external) of $\angle(T T)$ intersect the $x$-axis at points $E$ and $F$. The area of $\triangle TEF$ is $S$. Find:
(1) The coordinates of point $T$.
(2) The locus equation of the midpoint of $CD$ that minimizes $S$.
14. (1) Let $T\left(x_{0}, y_{0}\right)$, then $l: \frac{x_{0} x}{u^{2}}-\frac{y_{0} y}{b^{2}}=1$
(1.
The asymptotes of $C_{1}$ are $y= \pm \frac{b}{a}, d$
Substituting into (D) yields $x=\left(\frac{x_{0}}{a^{2}}-\frac{y_{n}}{a b}\right)^{1}{ }^{1} \cdot j x-\left(\frac{x_{0}}{a^{2}}+\frac{y_{0}}{a b}\right)^{-1}$
Thus, by $M$ being the midpoint of $A B$, we have:
$$
\begin{array}{l}
4=\left(\frac{r_{0}}{a^{2}}-\frac{y_{0}}{a b}\right)^{-1}+\left(\frac{x_{0}}{a^{2}}-\frac{y_{0}}{a b}\right)^{-1} \\
=\frac{2 a^{2} b^{2} \cdot c_{0}}{b^{2} x_{0}^{2}-a^{2} y_{0}^{2}}
\end{array}
$$

Since $\frac{x_{0}^{2}}{a^{2}}-\frac{y_{0}^{2}}{b^{2}}-1$, we have $2 x_{0}=4$, $x_{0}=2$.

Substituting $x= \pm \frac{b}{a} y$ into (1) yields $y_{0}=1$.
Thus, $T(2,1)$.
(2) Since $T E$ and $T F$ are the internal and external angle bisectors of $\angle C T D$,
we have $\angle E T F=\frac{\pi}{2}$ (assuming $E$ is to the left of $F$).
Construct $T H \perp E F$ at $H$, then $H(2,0)$.
Let $\angle E T H=\alpha$, so $\alpha \in\left(0, \frac{\pi}{2}\right)$.
At this point,
$$
\begin{aligned}
S_{\triangle \mathrm{TFF}} & =\frac{1}{2} T E \times T F \\
& =\frac{1}{2} \frac{T H}{\cos \alpha} \cdot \frac{T H}{\cos \left(\frac{\pi}{2}-\alpha\right)}
\end{aligned}
$$
$=\frac{1}{\sin ^{2} a} \geqslant 1$ with equality if and only if $\alpha=\frac{\pi}{4}$, i.e., $E(1,0)$ and $F(3,0)$.
On the other hand, consider the ellipse $\frac{x^{2}}{4}+y^{2}=1$, for which $C$ is $(0,1)$ and $D$ is $(2,0)$.
The points $E$ and $F$ determined by it are precisely $(1,0)$ and $(3,0)$, so the minimum value of $S$ is 1.
For all ellipses that make $E$ and $F$ be $(1,0)$ and $(3,0)$, $D$ lies on the segment $E F$ (since $O$ and $T$ are on $C D$). Let $\tan \angle D T E=k$, then $k \in(0,+\infty)$.
On the other hand, $\tan \angle E T C=k$.
Since $C$ is on the positive $y$-axis, we have $k>\tan \angle O T E=\frac{1}{3}$, so $k \in\left(\frac{1}{3},+\infty\right)$. When $D \neq H$, the slope of $T D$ exists. Let it be $k_{1}$.
Noting that the line $T D$ is symmetric to $T C$ about $T E$, and the slope of $T E$ is 1,
the slope of $T C$ is $\frac{1}{k_{1}}$;
$$
\begin{array}{l}
\tan \angle D T E=\frac{k_{1}-1}{1+k_{1}}=k ; \\
k_{1}=\frac{1+k}{1-k} .
\end{array}
$$

Since $T D: y-1=k_{1}(x-2)$, the $x$-coordinate of $D$ satisfies $y-1=\frac{1+k}{1-k}(x-2)$;
$$
D\left(\frac{3 k+1}{k+1}, 1\right) \text {. }
$$
$T C: y-1=\frac{1}{k_{1}}(x-2)$. So the $y$-coordinate of $C$ satisfies $y-1=\frac{1-k}{1+k} x-2$;
$$
\text { C }\left(0, \frac{3 k-1}{k+1}\right) \text {. }
$$

Thus, the midpoint $N$ of (D) is $\left(\frac{3 k+1}{2(k+1)}, \frac{3 k}{2(k+1)}\right)$.
Let $N(x, y)$. Then $x, y=\frac{3 k}{k+1}, y-\frac{1}{k+1}$.
So $k=\frac{x+y}{3-x}, \text{ and } k=\frac{1-x}{y-y}$.
Thus, $\frac{x+y}{3-x-y}-\frac{1-x+y}{x-y}$, i.e., $4 x-2 y \quad 3-\cdots$.
Noting that $k \in\left(\frac{1}{3},+\infty\right)$, we have $x=\frac{3 k+1}{3(k+1)}$, so $x \in\left(\frac{4}{3}, \frac{3}{2}\right)$.
Therefore, the equation of the trajectory of the midpoint of (D) is $4 x-2 y \quad 3-0, x \in\left(\frac{4}{3}, \frac{3}{2}\right)$.