Find the equation of the hyperbola satisfying the give conditions: Foci (0, ±13),

Foci $(0, \pm 13)$, the conjugate axis is of length 24 .

Here, the foci are on the $y$-axis.

Therefore, the equation of the hyperbola is of the form $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$.

Since the foci are $(0, \pm 13), c=13$.

Since the length of the conjugate axis is $24,2 b=24 \Rightarrow b=12$.

We know that $a^{2}+b^{2}=c^{2}$.

$\therefore a^{2}+12^{2}=13^{2}$

$\Rightarrow a^{2}=169-144=25$

Thus, the equation of the hyperbola is $\frac{y^{2}}{25}-\frac{x^{2}}{144}=1$.