17 (12 points) Given the ellipse $T: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ and the hyperbola $S$ : $\frac{x^{2}}{m^{2}}-\frac{y^{2}}{n^{2}}=1(m>0, n>0)$ have the same focus $F(2,0)$. Let the asymptote of the hyperbola $S$ in the first quadrant be $l$. If the focus $F$ and the upper vertex $B$ of the ellipse $T$ are symmetric with respect to the line $l$ and their symmetric points are on the hyperbola $S$, find the equations of the ellipse $T$ and the hyperbola $S$.
17 Given
$$
a^{2}-b^{2}=m^{2}+n^{2}=4,
$$

the equation of the asymptote $l$ is $y=\frac{n}{m} x$, and the equation of the line $l^{\prime}$, which passes through the focus $F$ and is perpendicular to the line $l$, is $y=-\frac{m}{n}(x-2)$. From $\left\{\begin{array}{l}y=\frac{n}{m} x, \\ y=-\frac{m}{n}(x-2),\end{array}\right.$ we get $\left\{\begin{array}{l}x=\frac{2 m^{2}}{m^{2}+n^{2}}, \\ y=\frac{2 m n}{m^{2}+n^{2}} .\end{array}\right.$ Since $m^{2}+n^{2}=4$, the lines $l$ and $l^{\prime}$ intersect at the point $\left(\frac{m^{2}}{2}, \frac{m n}{2}\right)$. Therefore, the symmetric point of the focus $F$ with respect to the line $l$ is $F^{\prime}\left(m^{2}-2, m n\right)$.
Given that the point $F^{\prime}$ lies on the hyperbola $S$, we have
$$
\frac{\left(m^{2}-2\right)^{2}}{m^{2}}-\frac{m^{2} n^{2}}{n^{2}}=1 \text {. }
$$

Solving this, we get $m^{2}=\frac{4}{5}, n^{2}=4-m^{2}=4-\frac{4}{5}=\frac{16}{5}$. Therefore, the equation of the hyperbola $S$ is $\frac{5 x^{2}}{4}-\frac{5 y^{2}}{16}=1$.

The coordinates of the upper vertex $B$ of the ellipse $T$ are $(0, b)$. Since $m^{2}=\frac{4}{5}, n^{2}=\frac{16}{5}$, the equation of the asymptote $l$ is $y=2 x$, and the equation of the line $l^{\prime \prime}$, which passes through the point $B$ and is perpendicular to the line $l$, is $y=-\frac{1}{2} x+b$. From $\left\{\begin{array}{l}y=2 x, \\ y=-\frac{1}{2} x+b .\end{array}\right.$ we get $\left\{\begin{array}{l}x=\frac{2}{5} b, \\ y=\frac{4}{5} b .\end{array}\right.$ Thus, the lines $l$ and $l^{\prime \prime}$ intersect at the point $\left(\frac{2}{5} b, \frac{4}{5} b\right)$. Therefore, the coordinates of the symmetric point of the point $B$ with respect to the line $l$ are $B^{\prime}\left(\frac{4}{5} b, \frac{3}{5} b\right)$.
Given that the point $B^{\prime}$ lies on the hyperbola $S$, we have
$$
\frac{5 \cdot\left(\frac{4}{5} b\right)^{2}}{4}-\frac{5 \cdot\left(\frac{3}{5} b\right)^{2}}{16}=1,
$$

Solving this, we get $b^{2}=\frac{16}{11}, a^{2}=b^{2}+4=\frac{60}{11}$, so the equation of the ellipse $T$ is $\frac{11 x^{2}}{60}+\frac{11 y^{2}}{16}=1$.