Solution: It is easy to know that the two foci of the ellipse are $F_{1}(-3,0)$, $F_{2}(3,0)$, and the lengths of the semi-major axis and semi-minor axis are $a$ and $b$ respectively,
$\because$ the line is tangent to the ellipse,
$\therefore d_{1} \cdot d_{2}=\frac{|-3+9|}{\sqrt{2}} \cdot \frac{|3+9|}{\sqrt{2}}=36=b^{2}$, $a^{2}=b^{2}+c^{2}=45$,
$\therefore$ the equation of the required ellipse is $\frac{x^{2}}{45}+\frac{y^{2}}{36}=1$.