Question:
Find the equation of the hyperbola satisfying the give conditions: Foci $(0, \pm 13)$, the conjugate axis is of length 24 .
Solution:
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 $\frac{y^{2}}{25}-\frac{x^{2}}{144}=1$.
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