# How do you write the equation of a hyperbola given center at the origin (0,0) and satisfies the given conditions: Foci F(0, -5) (0,5) and conjugate axis of length 4?

Jul 23, 2018

The equation is ${x}^{2} / 21 - {y}^{2} / 4 = 1$

#### Explanation:

The general equation of a hyperbola is

${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$

The center is $C = \left(h , k\right) = \left(0 , 0\right)$

The equation reduces to

${x}^{2} / {a}^{2} - {y}^{2} / {b}^{2} = 1$

The conjugate axis is

$2 b = 4$, $\implies$, $b = 2$

The equation is

${x}^{2} / {a}^{2} - {y}^{2} / 4 = 1$

The foci are $F = \left(0 , c\right) = \left(0 , 5\right)$ and $F ' = \left(0 , - c\right) = \left(0 , - 5\right)$

And

${c}^{2} = {a}^{2} + {b}^{2}$

Therefore,

${a}^{2} = {c}^{2} - {b}^{2} = {5}^{2} - 4 = 21$

Therefore,

The equation is

${x}^{2} / 21 - {y}^{2} / 4 = 1$

graph{x^2/21-y^2/4=1 [-18.02, 18.02, -9.01, 9.02]}