# How do you find the vertices and foci of x^2/121-y^2/4=1?

Nov 21, 2016

#### Explanation:

Standard form for a hyperbola with a horizontal transverse axis is:

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

Center: $\left(h , k\right)$
Vertices: $\left(h - a , k\right) \mathmr{and} \left(h + a , k\right)$
Foci: $\left(h - \sqrt{{a}^{2} + {b}^{2}} , k\right) \mathmr{and} \left(h + \sqrt{{a}^{2} + {b}^{2}} , k\right)$

Put the given equation in standard form:

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

vertices: $\left(- 11 , 0\right) \mathmr{and} \left(11 , 0\right)$

$\sqrt{{a}^{2} + {b}^{2}} = \sqrt{{11}^{2} + {2}^{2}} = \sqrt{125} = 5 \sqrt{5}$

Foci: $\left(- 5 \sqrt{5} , 0\right) \mathmr{and} \left(5 \sqrt{5} , 0\right)$