# How do you write an equation for the hyperbola with center (0,0), vertex (-2,0) and focus (4,0)?

Oct 30, 2016

#### Explanation:

The given, center, vertex, and focus share the same y coordinate, 0, ,therefore, the standard form for the equation of this type of hyperbola is the one corresponding to the Horizontal Transverse Axis type:

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

where $\left(h , k\right)$ is the center, $a$ is distance from the center to the vertex, and $b$ affects the distance, $c$, from the center to the focus as determined by the equation ${c}^{2} = {a}^{2} + {b}^{2}$.

Substitute the center $\left(0 , 0\right)$ into the standard form:

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

We know that $a = 2$, because one of the vertices a distance of 2 from the center. Substitute 2 for a into the equation:

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

Observing that the focus is a distance of 4 from the center we set ${c}^{2} = 16$ in the equation ${c}^{2} = {a}^{2} + {b}^{2}$:

$16 = {a}^{2} + {b}^{2}$

Substitute 4 for ${a}^{2}$

$16 = 4 + {b}^{2}$

Solve for b:

$b = \sqrt{12}$

Substitute $\sqrt{12}$ for b into the equation:

${\left(x - 0\right)}^{2} / {2}^{2} - {\left(y - 0\right)}^{2} / {\left(\sqrt{12}\right)}^{2} = 1$