# How do you find the vertices, asymptote, foci and graph x^2/64-(9y^2)/4=1?

Dec 6, 2016

#### Explanation:

Standard form for a Hyperbola of this type (Horizontal Transverse Axis) is:

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

The center is at: $\left(h , k\right)$
The vertices are at: $\left(h - a , k\right) \mathmr{and} \left(h + a , k\right)$
The foci are at: $\left(h - \sqrt{{a}^{2} + {b}^{2}} , k\right) \mathmr{and} \left(h + \sqrt{{a}^{2} + {b}^{2}} , k\right)$
The asymptotes are: $y = - \frac{b}{a} \left(x - h\right) + k \mathmr{and} y = \frac{b}{a} \left(x - h\right) + k$

Let put the given equation in standard form:

${\left(x - 0\right)}^{2} / {8}^{2} - {\left(y - 0\right)}^{2} / {2}^{2} = 1$

The center is at: $\left(0 , 0\right)$
The vertices are at: $\left(- 8 , 0\right) \mathmr{and} \left(8 , 0\right)$
The foci are at: $\left(- \sqrt{68} , 0\right) \mathmr{and} \left(\sqrt{68} , 0\right)$
The asymptotes are: $y = - \frac{1}{4} x \mathmr{and} y = \frac{1}{4} x$

Graph: