10. Through a point $P$ on the line $l: y=x+9$, construct an ellipse with the shortest major axis, whose foci are $F_{1}(-3,0)$ and $F_{2}(3,0)$. Then the equation of the ellipse is $\qquad$.
10. $\frac{x^{2}}{45}+\frac{y^{2}}{36}=1$.

As shown in Figure 4, let the point on line $l$ be $P(t, t+9)$, and take the symmetric point of $F_{1}(-3,0)$ about line $l$ as $Q(-9, 6)$. According to the definition of an ellipse, we have
$$
\begin{array}{l}
2 a=P F_{1}+P F_{2} \\
=P Q+P F_{2} \\
\geqslant Q F_{2} \\
=\sqrt{12^{2}+6^{2}}=6 \sqrt{5},
\end{array}
$$

The equality holds if and only if $Q$, $P$, and $F_{2}$ are collinear, i.e., $k_{P F_{2}}=k_{Q P_{2}}$, which means $\frac{t+9}{t-3}=\frac{6}{-12}$. 
At this point, $t=-5, P(-5,4)$.
Given $c=3, a=3 \sqrt{5}$, we get $a^{2}=45, b^{2}=36$.
Therefore, the equation of the ellipse is $\frac{x^{2}}{45}+\frac{y^{2}}{36}=1$.