Question

Find the equation of the hyperbola satisfying the given conditions.
Foci $(0,\pm 13),$ the conjugate axis is of lenth 24.

Answer 1

Here foci are $(0,\pm 13)$ which lie on y- axis.
So the equation of hyperbola in standard form is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$\therefore foci(0,\pm c)is(0,\pm 13)\Rightarrow c=13\text{length of conjugate axis}2b=24\Rightarrow b=12\text{We know that}{c}^2=a^2+b^2\therefore (13{)}^2=a^2+(12{)}^2\Rightarrow a^2=169-144=a^2=25$
Thus required equation of hyperbola is,
$\frac{y^2}{25}-\frac{x^2}{(12{)}^2}=1\Rightarrow \frac{y^2}{25}-\frac{x^2}{144}=1.$