Question

Find the equation of the hyperbola satisfying the give conditions: Foci $(0,\pm 13)$ the conjugate axis is of length $24$

Answer 1

It is given that, 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 $(0,\pm 13)\Rightarrow ae=c=13$
Since the length of the conjugate axis is $24$,$\Rightarrow 2b=24\Rightarrow b=12$
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 $\frac{y^2}{25}-\frac{x^2}{144}=1$