Given a line segment $|AB|=2a$, a moving point $P$ has a right angle view of $AB$, a moving point $Q$ is equidistant from $A$ and $B$, and $\triangle QAB$ has the same area as $\triangle PAB$ (but not zero). Point $O$ is the midpoint of $AB$, and $OP$ intersects $BQ$ at $M$. Taking $O$ as the origin and the perpendicular bisector of $AB$ as the $y$-axis, establish a coordinate system, and let $\angle BOP=\theta$. Use $\theta$ as a parameter to find the parametric equations of the trajectories of the moving points $P$, $Q$, and $M$, and convert these parametric equations into ordinary equations. Identify what these three trajectories are.

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Given a line segment $|AB|=2a$, a moving point $P$ has a right angle view of $AB$, a moving point $Q$ is equidistant from $A$ and $B$, and $\triangle QAB$ has the same area as $\triangle PAB$ (but not zero). Point $O$ is the midpoint of $AB$, and $OP$ intersects $BQ$ at $M$. Taking $O$ as the origin and the perpendicular bisector of $AB$ as the $y$-axis, establish a coordinate system, and let $\angle BOP=\theta$. Use $\theta$ as a parameter to find the parametric equations of the trajectories of the moving points $P$, $Q$, and $M$, and convert these parametric equations into ordinary equations. Identify what these three trajectories are.
(1) Let the coordinates of $P$ be $(x, y)$, then the parametric equations of the trajectory of point $P$ are $\left\{\begin{array}{l}x=a \cos \theta, \\ y=a \sin \theta\end{array}\right.$. $0<\theta<\pi$ or $\pi<\theta<2 \theta$. Eliminating $\theta$ yields $x^{2}+y^{2}=a^{2}(y \neq 0)$. Therefore, the trajectory of the moving point $P$ is a circle centered at the origin with radius $a$, excluding points $A$ and $B$.
(2) Let the coordinates of point $Q$ be $(x, y)$. Since $|Q A|=|Q B|$, and $\triangle Q A B$ has the same area as $\triangle P A B$, point $Q$ lies on the $y$-axis, with the same $y$-coordinate as $P$. Thus, the parametric equations are $\left\{\begin{array}{l}x=0, \\ y=a \sin \theta,\end{array} \quad(0<\theta<\pi\right.$ or $\pi<\theta<2 \theta)$. The ordinary equation is $x=0(|y| \leqslant a$ and $y \neq 0)$. Hence, the trajectory of $Q$ is the diameter of the circle on the $y$-axis, excluding point $O$.
(3) Let the coordinates of point $M$ be $(x, y)$. The equation of $O P$ is $y=x \tan \theta$, and the equation of $B Q$ is $y=(a-x) \sin \theta$. Solving these two equations yields $\left\{\begin{array}{l}x=\frac{a \cos \theta}{1+\cos \theta}, \\ y=\frac{a \sin \theta}{1+\cos \theta},\end{array}\right.$ Eliminating $\theta$ gives $y^{2}=-2 a\left(x-\frac{a}{2}\right)$. Therefore, the trajectory of point $M$ is the parabola $y^{2}=-2 a\left(x-\frac{a}{2}\right)$, excluding the vertex $\left(\frac{a}{2}, 0\right)$.