Four. (20 points) Let $A\left(x_{0}, y_{0}\right)$ be a point on the ellipse $C_{1}$:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$

Points $B$, $C$, and $D$ are the reflections of point $A$ across the $y$-axis, the origin, and the $x$-axis, respectively. Point $E$ is a point on the ellipse $C_{1}$ such that $A E \perp A C$, and $B D$ intersects $C E$ at point $F$.
(1) Find the equation of the trajectory $C_{2}$ of point $F$;
(2) Let $C_{3}: x^{2}+y^{2}=1$, and
$$
\left(a^{2}+b^{2}\right)^{3}=a^{2} b^{2}\left(a^{2}-b^{2}\right)^{2},
$$
A chord $A_{1} B_{1}$ of $C_{2}$ is tangent to $C_{3}$. Determine the measure of $\angle A_{1} O B_{1}$.
(1) From $A\left(x_{0}, y_{0}\right)$, we have
$$
B\left(-x_{0}, y_{0}\right), C\left(-x_{0},-y_{0}\right), D\left(x_{0},-y_{0}\right).
$$

Then $l_{B D}: y=-\frac{y_{0}}{x_{0}} x$.
Let $E\left(x_{1}, y_{1}\right)$. Since points $A$ and $E$ are both on the ellipse $C_{1}$, we have
$\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}=1$,
$\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}=1$.
(2) - (1) gives
$\frac{1}{a^{2}}\left(x_{1}+x_{0}\right)\left(x_{1}-x_{0}\right)+\frac{1}{b^{2}}\left(y_{1}+y_{0}\right)\left(y_{1}-y_{0}\right)=0$.
Clearly, $x_{1} \neq x_{0}$, thus,
$\frac{b^{2}}{a^{2}}+\frac{y_{1}-y_{0}}{x_{1}-x_{0}} \cdot \frac{y_{1}+y_{0}}{x_{1}+x_{0}}=0$.
Since $A C \perp A E$, we have $k_{A C} k_{A E}=-1$.
Then $k_{A E}=\frac{y_{1}-y_{0}}{x_{1}-x_{0}}=-\frac{1}{k_{A C}}=-\frac{x_{0}}{y_{0}}$.
Thus, $k_{C E}=\frac{y_{1}-\left(-y_{0}\right)}{x_{1}-\left(-x_{0}\right)}=\frac{y_{1}+y_{0}}{x_{1}+x_{0}}$.
From equation (3), we get $k_{C E}=\frac{b^{2} y_{0}}{a^{2} x_{0}}$.
By combining the equations of $l_{C E}: y+y_{0}=\frac{b^{2} y_{0}}{a^{2} x_{0}}\left(x+x_{0}\right)$ and $l_{B D}$, we solve for
$x_{F}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}} x_{0}, y_{F}=-\frac{a^{2}-b^{2}}{a^{2}+b^{2}} y_{0}$.
Since $x_{0} \neq 0, y_{0} \neq 0$, we have $x_{F} \neq 0, y_{F} \neq 0$.
From equation (1), the trajectory equation $C_{2}$ of point $F$ is
$\frac{x^{2}}{\frac{\left(a^{2}-b^{2}\right)^{2} a^{2}}{\left(a^{2}+b^{2}\right)^{2}}}+\frac{y^{2}}{\frac{\left(a^{2}-b^{2}\right)^{2} b^{2}}{\left(a^{2}+b^{2}\right)^{2}}}=1(x \neq 0, y \neq 0)$.
(2) From (1), let $a_{1}^{2}=\frac{\left(a^{2}-b^{2}\right)^{2} a^{2}}{\left(a^{2}+b^{2}\right)^{2}}$,
$b_{1}^{2}=\frac{\left(a^{2}-b^{2}\right)^{2} b^{2}}{\left(a^{2}+b^{2}\right)^{2}}\left(a_{1}>b_{1}>0\right)$.
Since $\left(a^{2}+b^{2}\right)^{3}=a^{2} b^{2}\left(a^{2}-b^{2}\right)^{2}$, we have
$\frac{1}{a_{1}^{2}}+\frac{1}{b_{1}^{2}}=\frac{\left(a^{2}+b^{2}\right)^{2}\left(a^{2}+b^{2}\right)}{\left(a^{2}-b^{2}\right)^{2} a^{2} b^{2}}=1$.
Thus, $C_{2}: \frac{x^{2}}{a_{1}^{2}}+\frac{y^{2}}{b_{1}^{2}}=1, C_{3}: x^{2}+y^{2}=1$, and $\frac{1}{a_{1}^{2}}+\frac{1}{b_{1}^{2}}=1$.
If $A_{1} B_{1} \perp x$-axis, then $l_{A_{1} B_{1}}: x=1$. We can calculate $y_{A_{1}}=1, y_{B_{1}}=-1$, thus $\angle A_{1} O B_{1}=90^{\circ}\left(l_{A_{1} B_{1}}: x=-1\right.$, we get the same conclusion).

If the slope of line $A_{1} B_{1}$ is $k$, let $l_{A_{1} B_{1}}: y=k x+$ $(m \neq 0)$. Since $A_{1} B_{1}$ is tangent to $\odot O$, then
$$
1=r=\frac{|m|}{\sqrt{1+k^{2}}},
$$

which means $m^{2}=k^{2}+1$.
Substituting $1=\frac{y-k x}{m}$ into $C_{2}$, we get
$$
\frac{x^{2}}{a_{1}^{2}}+\frac{y^{2}}{b_{1}^{2}}=\left(\frac{y-k x}{m}\right)^{2}.
$$

Converting the above equation to one with $\frac{y}{x}$ as the main variable, we get
$$
\left(\frac{1}{m^{2}}-\frac{1}{b_{1}^{2}}\right)\left(\frac{y}{x}\right)^{2}-\frac{2 k}{m^{2}}\left(\frac{y}{x}\right)+\frac{k^{2}}{m^{2}}-\frac{1}{a_{1}^{2}}=0.
$$

Since there are two intersection points, $m^{2} \neq b_{1}^{2}$.
By Vieta's formulas, we get
$$
\begin{array}{c}
\frac{y_{2}}{x_{2}} \cdot \frac{y_{3}}{x_{3}}=\frac{\frac{k^{2}}{m^{2}}-\frac{1}{a_{1}^{2}}}{\frac{1}{m^{2}}-\frac{1}{b_{1}^{2}}}=\frac{\frac{m^{2}-1}{m^{2}}-\frac{1}{a_{1}^{2}}}{\frac{1}{m^{2}}-\frac{1}{b_{1}^{2}}} \\
=\frac{1-\frac{1}{a_{1}^{2}}-\frac{1}{m^{2}}}{\frac{1}{m^{2}}-\frac{1}{b_{1}^{2}}}=\frac{\frac{1}{b_{1}^{2}}-\frac{1}{m^{2}}}{\frac{1}{m^{2}}-\frac{1}{b_{1}^{2}}}=-1,
\end{array}
$$

where $A_{1}\left(x_{2}, y_{2}\right)$ and $B_{1}\left(x_{3}, y_{3}\right)$.
Therefore, $A_{1} O \perp B_{1} O$, and $\angle A_{1} O B_{1}=90^{\circ}$.