10. As shown in Figure $3, A$ is a point on the positive half-axis of $x$, $\angle A O N=60^{\circ} ; B$ is a point on $O N$, and $A B=2$; $C$ is the midpoint of $A B$, and its symmetric point about $O B$ is $D$. Find the equation of the trajectory of point $D$, and draw a rough graph.
10. According to the problem, let point $A(2 a, 0)\left(0 \leqslant a \leqslant \frac{2 \sqrt{3}}{3}\right)$, point $B(b, \sqrt{3} b)\left(0 \leqslant b \leqslant \frac{2 \sqrt{3}}{3}\right)$.
Then the symmetric point of point $A$ with respect to $O B$ is $E(-a, \sqrt{3} a)$. Let point $D(x, y)$.
Since $D$ is the midpoint of $B E$, we have,
$$
D\left(\frac{b-a}{2}, \frac{\sqrt{3}}{2}(a+b)\right) \text {, }
$$

i.e., $\square$ $x=\frac{b-a}{2}, y=\frac{\sqrt{3}}{2}(a+b)$.
From $(b-2 a)^{2}+(\sqrt{3} b)^{2}=4$
$\Rightarrow a^{2}+b^{2}-a b=1$
$\Rightarrow(a+b)^{2}+3(b-a)^{2}=4$
$\Rightarrow 3 x^{2}+\frac{y^{2}}{3}=1\left(-\frac{1}{2} \leqslant x \leqslant \frac{1}{2}, \frac{\sqrt{3}}{2} \leqslant y \leqslant \sqrt{3}\right)$.
Therefore, the equation of the trajectory of point $D$ is
$$
3 x^{2}+\frac{y^{2}}{3}=1\left(-\frac{1}{2} \leqslant x \leqslant \frac{1}{2}, \frac{\sqrt{3}}{2} \leqslant y \leqslant \sqrt{3}\right),
$$

The rough graph of its trajectory is shown in Figure 5.