# How do you write the equation for a hyperbola given vertices (0,-4) and (0,4), conjugate axis of lengths 14 units?

Jun 30, 2017

Equation of hyperbola is ${y}^{2} / 16 - {x}^{2} / 49 = 1$

#### Explanation:

As the vertices are $\left(0 , - 4\right)$ and $\left(0 , 4\right)$ i.e. along $y$-axis, we have a vertical hyperbola and equation of hyperbola is of the type

${\left(y - k\right)}^{2} / {a}^{2} - {\left(x - h\right)}^{2} / {b}^{2} = 1$, where $\left(h , k\right)$ is center of hyperbola and $b > a$

As both vertices are equidistant from origin, center of hyperbola is $\left(0 , 0\right)$ and the equation is

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

Further as length of conjugate axis is $14$, we have $b = \frac{14}{2} - 7$

and hence equation of hyperbola is ${y}^{2} / 16 - {x}^{2} / 49 = 1$
graph{y^2/16-x^2/49=1 [-20, 20, -10, 10]}