9. The line $l$ intersects the circle $x^{2}+y^{2}=a^{2}$ at points $P, Q$, and intersects the hyperbola $x^{2}-y^{2}=a^{2}$ at points $M, N$, such that $P, Q$ trisect the segment $M N$. Find the equation of the line $l$.
9. Solution: Let the coordinates of $P, Q, M, N$ be $\left(x_{i}, y_{i}\right), i=1,2,3,4, l: y=m x+b$. From $\left\{\begin{array}{l}y=m x+b \\ x^{2}+y^{2}=a^{2}\end{array}\right.$ we get
$$
\left(1+m^{2}\right) x^{2}+2 m b x+b^{2}-a^{2}=0, \therefore x_{1}+x_{2}=-\frac{2 m b}{1+m^{2}},
$$

Similarly, $x_{3}, x_{4}$ are the roots of the equation $\left(1-m^{2}\right) x^{2}-2 m b x-\left(b^{2}+a^{2}\right)=0$, i.e., $x_{3}+x_{4}=\frac{2 m b}{1-m^{2}}$,
$\therefore-\frac{2 m b}{1+m^{2}}=\frac{2 m b}{1-m^{2}}$, solving this gives $m=0$ or $b=0$.
(1) If $m=0$, then the equation becomes $x^{2}+b^{2}-a^{2}=0, x^{2}-b^{2}-a^{2}=0$,
$\therefore x_{1,2}= \pm \sqrt{a^{2}-b^{2}}, x_{3,4}= \pm \sqrt{a^{2}+b^{2}}$, and from $|P Q|=\frac{1}{3}|M N|$ we get $\left|x_{1}-x_{2}\right|=\frac{1}{3}\left|x_{3}-x_{4}\right|$, i.e., $2 \sqrt{a^{2}-b^{2}}=\frac{2}{3} \sqrt{a^{2}+b^{2}}$, solving this gives $b= \pm \frac{2 \sqrt{5}}{5} a$.
(2) When $b=0$, similarly we get $m= \pm \frac{2 \sqrt{5}}{5}$, in summary, the equation of $l$ is $y= \pm \frac{2 \sqrt{5}}{5} x$ or $y= \pm \frac{2 \sqrt{5}}{5} a$.