13.3.9 * ћ Two rods rotate around points $A(a, 0)$ and $B(-a, 0)$, and the product of their intercepts on the $y$-axis $b b_{1}=a^{2}$ (constant). Try to find: the equation of the trajectory of the intersection point of the rotating rods.
Let $M(x, y)$ be the intersection point of the two rods, then $M(x, y)$ satisfies the linear equations defined by the two rods $y=-\frac{b}{a}(x-a)$,
(1) $y=\frac{b_{1}}{a}(x+a)$,
(2) Multiplying both sides of (1) and (2), we get $y^{2}=-\frac{b b_{1}}{a^{2}}\left(x^{2}-a^{2}\right)$. Substituting $b b_{1}=a^{2}$ (given) into the above equation, we get $y^{2}=-x^{2}+a^{2}$, which simplifies to $x^{2}+y^{2}=a^{2}$. Since $b b_{1}=a^{2}$ (a constant), $b$ and $b_{1}$ are not zero, thus $y \neq 0, x \neq \pm a$. Therefore, the required trajectory equation is $x^{2}+y^{2}=a^{2}(-a<x<a)$.