2. Given the equation of $\odot O_{1}$ is $(x-8)^{2}+y^{2}=4$, and the equation of $\odot O_{2}$ is $(x-12)^{2}+y^{2}=9$, points $A$ and $B$ are on $\odot O_{1}$ and $\odot O_{2}$ respectively, and $A B$ is the external common tangent of the two circles. A moving line $l \parallel A B$ intersects $\odot O_{1}$ at point $X$ and $\odot O_{2}$ at point $Y$, and $A X$ intersects $B Y$ at point $P$. Then the locus of point $P$ is $\qquad$.
2. $x^{2}+y^{2}=90$ H $x \neq \frac{75}{8}$.
From 1 $O O_{1}=8$,
$O O_{2}=12$ we know $O, A$.
$B$. Points are collinear. Let $A B$
$\operatorname{l} O_{1} O_{2}$ intersect at point $O$,
with $O$ as the origin.
$O A$ as the $x$-axis, establish
a rectangular coordinate system as shown in Figure 6.
Then the equation of $\odot O_{1}$ is $(x-2 \sqrt{15})^{2}+(y-2)^{2}=4$,
the equation of $\odot O_{2}$ is $(x-3 \sqrt{15})^{2}+(y-3)^{2}=9$.
Let $Y\left(x_{1}, m\right), Y\left(x_{2}, m\right)$.
$X A(2 \sqrt{15}, 0), B(3 \sqrt{15}, 0)$, then
$x_{1}=2 \sqrt{15} \pm \sqrt{4 m-m^{2}}$,
$x_{2}=3 \sqrt{15} \pm \sqrt{6 m-m^{2}}$.
Thus, the equations of lines $A \mathrm{I}$ and $B Y$ are respectively
$$
\begin{array}{l}
y= \pm \frac{m}{\sqrt{4 m-m^{2}}}(x-2 \sqrt{15}), \\
y= \pm \frac{m}{\sqrt{6 m}-m^{2}}(x-3 \sqrt{15}) .
\end{array}
$$

Let $P\left(x_{0}, y_{0}\right)$. Then we have
$$
\begin{array}{c}
\left(\frac{x_{0}-2 \sqrt{15}}{y_{0}}\right)^{2}=\frac{4 m-m^{2}}{m^{2}} . \\
\left(\frac{x_{0}-3 \sqrt{15}}{y_{0}}\right)^{2}=\frac{6 m-m^{2}}{m^{2}} .
\end{array}
$$

Eliminating $m$ gives $x_{0}^{2}+y_{0}^{2}=90$. Therefore, the trajectory of point $P$ is a circle with $O$ as the center and $3 \sqrt{10}$ as the radius, and $P$ does not pass through the intersection points $C, D$ (where $x_{C}=x_{D}=\frac{75}{8}$).