18.4 Large Question: Let $A$ be a point on the ellipse $C_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$. Its symmetric points with respect to the $y$-axis, the origin, and the $x$-axis are $B$, $C$, and $D$ respectively. $E$ is another point on the ellipse $C_{1}$ such that $A E \perp A C$.
(1) Find the equation of the trajectory $C_{2}$ of the intersection point $F$ of the lines $B D$ and $C E$;
(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 $C_{3}$. Find the measure of $\angle A_{1} O B_{1}$.
(1) Let $A\left(x_{0}, y_{0}\right)$, then by symmetry, $B\left(-x_{0}, y_{0}\right), C\left(-x_{0},-y_{0}\right), D\left(x_{0},-y_{0}\right)$ and $x_{0} y_{0} \neq 0$, quadrilateral $A B C D$ is a rectangle. Let $E\left(x_{1}, y_{1}\right)$, since $A, E$ are both on the ellipse $C_{1}$, then
$$
\left\{\begin{array}{l}
b^{2} x_{0}^{2}+a^{2} y_{0}^{2}=a^{2} b^{2}, \\
b^{2} x_{1}^{2}+a^{2} y_{1}^{2}=a^{2} b^{2},
\end{array}\right.
$$

Subtracting (1) from (2) gives $b^{2}\left(x_{1}+x_{0}\right)\left(x_{1}-x_{0}\right)+a^{2}\left(y_{1}+\right.$
$\left.y_{0}\right)\left(y_{1}-y_{0}\right)=0$, clearly $x_{1} \neq x_{0}$, thus
$$
\frac{y_{1}+y_{0}}{x_{1}+x_{0}}=-\frac{b^{2}}{a^{2}} \frac{x_{1}-x_{0}}{y_{1}-y_{0}}=-\frac{b^{2}}{a^{2}} \frac{1}{k_{A E}}=\frac{b^{2}}{a^{2}} k_{A C}=\frac{b^{2}}{a^{2}} \frac{y_{0}}{x_{0}},
$$

where this is because $A E \perp A C$, so $k_{A E} \cdot k_{A C}=-1$. Therefore,
$$
\left\{\begin{array}{l}
l_{B D}: y=-\frac{y_{0}}{x_{0}} x, \\
l_{C E}: y+y_{0}=\frac{y_{1}+y_{0}}{x_{1}+x_{0}}\left(x+x_{0}\right)=\frac{b^{2}}{a^{2}} \frac{y_{0}}{x_{0}}\left(x+x_{0}\right),
\end{array}\right.
$$

Solving the system of equations, we get $\left\{\begin{array}{l}x_{0}=\frac{a^{2}+b^{2}}{a^{2}-b^{2}} x, \\ y_{0}=-\frac{a^{2}+b^{2}}{a^{2}-b^{2}} y,\end{array}\right.$ Substituting into the ellipse equation, and since $x_{0} y_{0} \neq 0$, we have $x y \neq 0$. Thus, the trajectory equation of intersection point $F$ is $C_{2}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\left(\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\right)^{2}(x y \neq 0)$.
(2) Let $p^{2}=\left(\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\right)^{2} a^{2}, q^{2}=\left(\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\right)^{2} b^{2} \quad$ and $p>q>0$, and since $\left(a^{2}+\right.$ $\left.b^{2}\right)^{3}-a^{2} b^{2}\left(a^{2}-b^{2}\right)^{2}$, we have $\frac{1}{p^{2}}+\frac{1}{q^{2}}=\left(\frac{a^{2}+b^{2}}{a^{2}-b^{2}}\right)^{2} \frac{a^{2}+b^{2}}{a^{2} b^{2}}=1$, hence $C_{2}: \frac{x^{2}}{p^{2}}+\frac{y^{2}}{q^{2}}=$ $1, C_{3}: x^{2}+y^{2}=1$, and $\frac{1}{p^{2}}+\frac{1}{q^{2}}=1$.

If $A_{1} B_{1} \perp x$-axis, by symmetry, assume $l_{A_{1} B_{1}}: x=1$, solving gives $y A_{1}=1, y B_{1}=-1$, so $\angle A_{1} O B_{1}=\frac{\pi}{2}$;

If $A_{1} B_{1} \chi x$-axis, let $l_{A_{1} B_{1}}: y=k x+m$, then $m \neq 0$. Since $A_{1} B_{1}$ is tangent to circle $C_{3}$, then $1=\frac{|m|}{\sqrt{1+k^{2}}}$, i.e., $m^{2}=k^{2}+1$. Substituting $1=\frac{y-k x}{m}$ into $C_{2}$, we get $\frac{x^{2}}{p^{2}}+\frac{y^{2}}{q^{2}}=1=$ $\left(\frac{y-k x}{m}\right)^{2}$. Rearranging into an equation with $\frac{y}{x}$ as the main variable gives
$$
\left(\frac{1}{m^{2}}-\frac{1}{q^{2}}\right)\left(\frac{y}{x}\right)^{2}-\frac{2 k}{m^{2}} \frac{y}{x}+\frac{k^{2}}{m^{2}}-\frac{1}{p^{2}}=0,
$$

Since there are two intersection points, $m^{2} \neq q^{2}$, let the two roots be $\frac{y_{2}}{x_{2}}$ and $\frac{y_{3}}{x_{3}}$, i.e., $A_{1}\left(x_{2}, y_{2}\right), B_{1}\left(x_{3}, y_{3}\right)$, then
$$
\frac{y_{2}}{x_{2}} \cdot \frac{y_{3}}{x_{3}}=\frac{\frac{k^{2}}{m^{2}}-\frac{1}{p^{2}}}{\frac{1}{m^{2}}-\frac{1}{q^{2}}}=\frac{\frac{m^{2}-1}{m^{2}}-\frac{1}{p^{2}}}{\frac{1}{m^{2}}-\frac{1}{q^{2}}}=\frac{\frac{1}{q^{2}}-\frac{1}{m^{2}}}{\frac{1}{m^{2}}-\frac{1}{q^{2}}}=-1,
$$

Thus, $A_{1} O \perp B_{1} O$, so $\angle A_{1} O B_{1}=\frac{\pi}{2}$. In summary, $\angle A_{1} O B_{1}=\frac{\pi}{2}$.