- Home
- Standard 11
- Mathematics
Find the equation of the hyperbola satisfying the give conditions: Foci $(0,\,\pm 13),$ the conjugate axis is of length $24.$
$\frac{y^{2}}{25}-\frac{x^{2}}{144}=1$
$\frac{y^{2}}{25}-\frac{x^{2}}{144}=1$
$\frac{y^{2}}{25}-\frac{x^{2}}{144}=1$
$\frac{y^{2}}{25}-\frac{x^{2}}{144}=1$
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 $\frac{y^{2}}{25}-\frac{x^{2}}{144}=1$
