13.S. Given that the ratio of the length of the tangent from point $P$ to the circle $(x+2)^{2}+y^{2}=1$ to the distance from $P$ to the $y$-axis is $\iota(\iota>0, \iota \neq 1)$. Let the locus of the moving point $P$ be $C$, and the two foci of curve $C$ be $F_{1}$ and $F_{2}$. If there is no point $Q$ on curve $C$ such that $\angle F_{1} Q F_{2}=\theta(0<\theta<\pi)$, where $\theta$ is given, find: the range of values of $t$ that satisfy the condition.
Let $P\left(x_{0}, y_{0}\right)$, $PM$ be the tangent line, $M$ be the point of tangency, $O$ be the origin, and the center of the circle $T(-2,0)$. Then
$$
|P M|^{2}=|P T|^{2}-r^{2}=\left(x_{0}+2\right)^{2}+y_{0}^{2}-1=t^{2} x_{0}^{2},
$$

which means the trajectory $C$ of point $P$ is
$$
\left(1-t^{2}\right) x^{2}+y^{2}+4 x+3=0,
$$

or
$$
\begin{array}{c}
\left(1-t^{2}\right)\left[x^{2}+\frac{4}{1-t^{2}} x+\frac{4}{\left(1-t^{2}\right)^{2}}\right]+y^{2}=-3+\frac{4}{1-t^{2}}, \\
\frac{\left(x+\frac{2}{1-t^{2}}\right)^{2}}{\frac{3 t^{2}+1}{\left(1-t^{2}\right)^{2}}}+\frac{y^{2}}{\frac{3 t^{2}+1}{1-t^{2}}}=1 .
\end{array}
$$
( I ) If $1-t^{2}>0$, i.e., $t^{2}<1$, and $1-t^{2}<1$, then the ellipse $C$ has a semi-major axis $a=\sqrt{\frac{3 t^{2}+1}{\left(1-t^{2}\right)^{2}}}$, a semi-minor axis $b=\sqrt{\frac{3 t^{2}+1}{1-t^{2}}}$, and a semi-focal distance $c=\sqrt{a^{2}-b^{2}}=\sqrt{\frac{\left(3 t^{2}+1\right) t^{2}}{\left(1-t^{2}\right)^{2}}} \Rightarrow \frac{c}{b}=\sqrt{\frac{t^{2}}{1-t^{2}}}$. Translating the center of the ellipse to the origin $O$, and for a point $Q$ on the ellipse, we have $S_{\triangle F_{1} F_{2}}=\frac{1}{2} \cdot$
$$
\begin{array}{l}
\left|F_{1} Q\right| \cdot\left|F_{2} Q\right| \cdot \sin \angle F_{1} Q F_{2} \text {, and } \\
\quad\left|F_{1} Q\right|^{2}+\left|F_{2} Q\right|^{2}-2\left|F_{1} Q\right| \cdot\left|F_{2} Q\right| \cdot \cos \angle F_{1} Q F_{2}=\left|F_{1} F_{2}\right|^{2}=4 c^{2},
\end{array}
$$

which means $\left(\left|F_{1} Q\right|+\left|F_{2} Q\right|\right)^{2}-2\left|F_{1} Q\right| \cdot\left|F_{2} Q\right|\left(1+\cos \angle F_{1} Q F_{2}\right)=4 c^{2}$, and $\left|F_{1} Q\right|+$ $\left|F_{2} Q\right|=2 a$, so
$$
\begin{array}{c}
\left|F_{1} Q\right| \cdot\left|F_{2} Q\right|=\frac{4 b^{2}}{2\left(1+\cos \angle F_{1} Q F_{2}\right)}, \\
S_{\triangle F_{1} Q F_{2}}=b^{2} \cdot \frac{\sin \angle F_{1} Q F_{2}}{1+\cos \angle F_{1} Q F_{2}}=b^{2} \cdot \tan \frac{\angle F_{1} Q F_{2}}{2},
\end{array}
$$

Since $S_{\triangle F_{1} Q F_{2}}$ is maximized when $Q$ is at the top vertex, we have
$$
\tan \frac{\angle F_{1} Q F_{2}}{2} \leqslant \frac{\frac{1}{2} \times 2 b c}{b^{2}}=\frac{c}{b}=\sqrt{\frac{t^{2}}{1-t^{2}}},
$$

By the problem's condition, we only need $\sqrt{\frac{t^{2}}{1-t^{2}}}<\tan \frac{\theta}{2}$, which implies $t^{2}<\frac{\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}} \Rightarrow t<\sin \frac{\theta}{2}$, so the range of $t$ is $0<t<\sin \frac{\theta}{2}$.