# How do you identify the vertices, foci, and direction of x^2/81-y^2/4=1?

Jan 2, 2017

#### Explanation:

The reference Conics: Hyperbolas tells us that the given equation:

${x}^{2} / 81 - {y}^{2} / 4 = 1 \text{ [1]}$

is that of a hyperbola with a horizontal transverse axis. We can identify that the direction is horizontal, because the "x" term is positive and the "y" term is negative.

NOTE: If the "y" term were positive and "x" term were negative, then the direction would be vertical.

The reference, also, tells us that the following equation, [2], is the standard Cartesian form:

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

In this form, it is easy to observe or compute:

1. The center is $\left(h , k\right)$
2. The vertices are located at $\left(h - a , k\right) \mathmr{and} \left(h + a , k\right)$
3. The foci are located at $\left(h - \sqrt{{a}^{2} + {b}^{2}} , k\right) \mathmr{and} \left(h + \sqrt{{a}^{2} + {b}^{2}} , k\right)$

Write equation [1] in the form of equation [2]:

${\left(x - 0\right)}^{2} / {9}^{2} - {\left(y - 0\right)}^{2} / {2}^{2} = 1 \text{ [3]}$

In this form, the vertices can be written by observation:

$\left(- 9 , 0\right)$ and $\left(9 , 0\right)$

Compute $\sqrt{{a}^{2} - {b}^{2}} = \sqrt{{9}^{2} + {2}^{2}} = \sqrt{85}$

The foci are:

$\left(- \sqrt{85} , 0\right)$ and $\left(\sqrt{85} , 0\right)$