$O B$ are two moving points, $O D \perp A B$ at $D$, $E$ is the midpoint of $A B$. Find the trajectory equation of $D$ and the trajectory equation of $E$.
Let the two trajectories be denoted as $C_{1}$ and $C_{2}$. For any point $P\left(x_{0}, y_{0}\right)$ on $C_{1}$, $P$ is inside the ellipse and $P$ is not the origin. Let the inclination angle of $OP$ be $\theta$. Then the parametric equation of $AB$ can be set as
$$
\left\{\begin{array}{l}
x=x_{0}+t \cos \left(\theta+\frac{\pi}{2}\right)=x_{0}-t \sin \theta, \\
y=y_{0}+t \sin \left(\theta+\frac{\pi}{2}\right)=y_{0}+t \cos \theta,
\end{array}\right.
$$

Substituting into the equation of the ellipse $C$ gives $\frac{\left(x_{0}-t \sin \theta\right)^{2}}{a^{2}}+\frac{\left(y_{0}+t \cos \theta\right)^{2}}{b^{2}}=1$. Simplifying, we get
$$
\left(\frac{\sin ^{2} \theta}{a^{2}}+\frac{\cos ^{2} \theta}{b^{2}}\right) t^{2}+2\left(\frac{y_{0} \cos \theta}{b^{2}}-\frac{x_{0} \sin \theta}{a^{2}}\right) t+\left(\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}-1\right)=0,
$$

Let the two roots be $t_{1}$ and $t_{2}$. Then $O A \perp O B \Leftrightarrow|P A| \cdot|P B|=|O P|^{2} \Leftrightarrow t_{1} t_{2}+x_{0}^{2}+y_{0}^{2}=0 \Leftrightarrow \frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}-1+\frac{1}{a^{2}} y_{0}^{2}+\frac{1}{b^{2}} x_{0}^{2}=0 \Leftrightarrow x_{0}^{2}+y_{0}^{2}=\frac{a^{2} b^{2}}{a^{2}+b^{2}}$, thus $C_{1}$ is the circle $x^{2}+y^{2}=\frac{a^{2} b^{2}}{a^{2}+b^{2}}$.

For any point $Q\left(x_{1}, y_{1}\right)$ on $C_{2}$, $Q$ is inside the ellipse and $Q$ is not the origin. Let the inclination angle of $AB$ be $\alpha$. Then the parametric equation of $AB$ is $l_{A B}:\left\{\begin{array}{l}x=x_{1}+t \cos \alpha, \\ y=y_{1}+t \sin \alpha,\end{array}\right.$ (parametric). Substituting into the equation of $C$, we get
$$
\left(b^{2} \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha\right) t^{2}+2\left(a^{2} y_{1} \sin \alpha+b^{2} x_{1} \cos \alpha\right) t+b^{2} x_{1}^{2}+a^{2} y_{1}^{2}-a^{2} b^{2}=0,
$$
$Q$ is the midpoint of $AB$ and
$$
\begin{aligned}
O A \perp O B & \Leftrightarrow|O Q|-|Q A|=|Q B| \\
& \Leftrightarrow\left\{\begin{array}{l}
a^{2} y_{1} \sin \alpha+b^{2} x_{1} \cos \alpha=0, \\
\left(b^{2} \cos ^{2} \alpha+a^{2} \sin ^{2} \alpha\right)\left(x_{1}^{2}+y_{1}^{2}\right)+b^{2} x_{1}^{2}+a^{2} y_{1}^{2}-a^{2} b^{2}=0,
\end{array}\right.
\end{aligned}
$$

From (1), we have $\left(a^{4} y_{1}^{2}+b^{4} x_{1}^{2}\right) \cos ^{2} \alpha=a^{4} y_{1}^{2}$. Substituting into (2), we get
$$
\left[a^{2}+\left(b^{2}-a^{2}\right) \frac{a^{4} y_{1}^{2}}{a^{4} y_{1}^{2}+b^{4} x_{1}^{2}}\right]\left(x_{1}^{2}+y_{1}^{2}\right)+b^{2} x_{1}^{2}+a^{2} y_{1}^{2}-a^{2} b^{2}=0,
$$

Simplifying, we get
$$
a^{2} b^{2}\left(\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}\right)\left(x_{1}^{2}+y_{1}^{2}\right)+\left(\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}-1\right)\left(a^{4} y_{1}^{2}+b^{4} x_{1}^{2}\right)=0,
$$

Thus, the equation of $C_{2}$ is
$$
a^{2} b^{2}\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\right)\left(x^{2}+y^{2}\right)+\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-1\right)\left(a^{4} y^{2}+b^{4} x^{2}\right)=0 \quad (x \neq 0).
$$

Therefore, the equation of $C_{1}$ is $x^{2}+y^{2}=\frac{a^{2} b^{2}}{a^{2}+b^{2}}$, and the equation of $C_{2}$ is $\left(a^{2}+b^{2}\right) a^{2} b^{2}\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\right)^{2}=$
$$
b^{4} x^{2}+a^{4} y^{2}
$$