# How do you identify the vertices, foci, and direction of x^2/121-y^2/36=1?

Apr 10, 2017

Given the equation of a hyperbola with a transverse horizontal axis:
${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$
The vertices are: $\left(h - a , k\right)$ and $\left(h + a , k\right)$
The foci are: $\left(h - \sqrt{{a}^{2} + {b}^{2}} , k\right)$ and $\left(h + \sqrt{{a}^{2} + {b}^{2}} , k\right)$

#### Explanation:

Note: A horizontal transverse type has a positive x and a negative y term, in a form that is equal to 1, and a vertical transverse type has a negative x term and a positive y term, in a form that is equal to 1.

Given: ${x}^{2} / 121 - {y}^{2} / 36 = 1$

Write in the standard form:

${\left(x - 0\right)}^{2} / {11}^{2} - {\left(y - 0\right)}^{2} / {6}^{2} = 1$

Now, we can write the vertices by observation:

$\left(- 11 , 0\right)$ and $\left(11 , 0\right)$

Compute $\sqrt{{a}^{2} + {b}^{2}} = \sqrt{{11}^{2} + {6}^{2}} = \sqrt{121 + 36} = \sqrt{157}$

Now, we can write the foci with the same ease:

$\left(- \sqrt{157} , 0\right)$ and $\left(\sqrt{157} , 0\right)$