5. If the two foci of the hyperbola $\Gamma_{1}$ are the two vertices of the ellipse $\Gamma_{2}: \frac{x^{2}}{5^{2}}+\frac{y^{2}}{4^{2}}=1$, and the two directrices of the hyperbola $\Gamma_{1}$ pass through the two foci of the ellipse $\Gamma_{2}$, then the equation of the hyperbola $\Gamma_{1}$ is $\qquad$ .
5. $\frac{x^{2}}{15}-\frac{y^{2}}{10}=1$

From the problem, we know that the vertices of the ellipse's major axis are $A(-5,0)$ and $B(5,0)$.
Thus, its foci are on the $x$-axis.
Let $c_{1}$ and $c_{2}$ represent the semi-focal distances of $\Gamma_{1}$ and $\Gamma_{2}$, respectively, then $c_{1}=5$, and
$$
c_{2}^{2}=5^{2}-4^{2}=9 \Rightarrow c_{2}=3 \text {. }
$$

Therefore, the foci of the ellipse are $F_{1}(-3,0)$ and $F_{2}(3,0)$, and the real axis of the hyperbola is the $x$-axis.
Let $\Gamma_{1}: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.
$$
\text { From } \frac{a^{2}}{c_{1}}=3 \Rightarrow a^{2}=15 \text {. }
$$

Thus, $b^{2}=c_{1}^{2}-a^{2}=5^{2}-15=10$.
Therefore, the equation of the hyperbola $\Gamma_{1}$ is $\frac{x^{2}}{15}-\frac{y^{2}}{10}=1$.