# How do you write an equation for the hyperbola with focus at (0, 25), asymptotes intersecting at (0, 5) and an asymptote passing through (24, 13)?

Aug 31, 2016

${\left(y - 5\right)}^{2} / 64 - {x}^{2} / 336 = 1$

#### Explanation:

The asymptotes intersect at the center $C \left(0. , 5\right)$.

A focus is $S \left(0 , 25\right)$.

$C S = a e = 25 - 5 = 20$.,

where a is the semi transverse axis and e is the eccentricity.

The transverse axis is along the y-axis. So, the equation of the

hyperbola is

${\left(y - 5\right)}^{2} / {a}^{2} - {\left(x - 0\right)}^{2} / \left({a}^{2} \left(1 - {e}^{2}\right)\right) = 1$

As it passes through (24, 13),

${8}^{2} / {a}^{2} - {24}^{2} / \left({a}^{2} \left({e}^{2} - 1\right)\right) = 1$

Substituting a = 20/e and simplifying,

$4 {e}^{4} - 29 {e}^{2} + 25 = 0$

$4 {e}^{4} - 25 {e}^{2} - 4 {e}^{2} + 25 = 0$

${e}^{2} \left(4 {e}^{2} - 25\right) - \left(4 {e}^{2} - 25\right) = 0$

$\left(4 {e}^{2} - 25\right) \left({e}^{2} - 1\right) = 0$

As e > 1,

$4 {e}^{2} - 25 = 0$

$e = \frac{5}{2}$.

$a = \frac{20}{e} = \frac{20}{\frac{5}{2}} = 8$
^Semi conjugate axis $b = a \sqrt{{e}^{2} - 1} = 8 \sqrt{\frac{21}{4}} \mathmr{and} {b}^{2} = 336$

And now, the equation is obtained as

${\left(y - 5\right)}^{2} / 64 - {x}^{2} / 336 = 1$-.