# How do you write the equation of the hyperbola given Foci: (-6,0),(6,0) and vertices (-5,0), (5,0)?

Jul 26, 2016

$\frac{11}{25} {x}^{2} - {y}^{2} - 11 = 0$

#### Explanation:

Calling $p = \left\{x , y\right\}$, ${f}_{1} = \left\{- 6 , 0\right\}$, ${f}_{2} = \left\{6 , 0\right\}$, ${v}_{1} = \left\{- 5 , 0\right\}$ and ${v}_{2} = \left\{5 , 0\right\}$ the hyperbole is deffined as the geometric place where

$\left\lVert p - {f}_{1} \right\rVert - \left\lVert p - {f}_{2} \right\rVert = \left\lVert {v}_{1} - {f}_{2} \right\rVert - \left\lVert {v}_{1} - {f}_{1} \right\rVert = \delta$

or

$\left\lVert p - {f}_{1} \right\rVert = \delta + \left\lVert p - {f}_{2} \right\rVert$

Squaring both sides

$\left\langlep - {f}_{1} , p - {f}_{1}\right\rangle = \left\langlep - {f}_{2} , p - {f}_{2}\right\rangle + 2 \delta \left\lVert p - {f}_{2} \right\rVert + {\delta}^{2}$

or

$\left\langlep - {f}_{1} , p - {f}_{1}\right\rangle - \left\langlep - {f}_{2} , p - {f}_{2}\right\rangle - {\delta}^{2} = 2 \delta \left\lVert p - {f}_{2} \right\rVert$

simplifying

$\left\langle{f}_{1} , {f}_{1}\right\rangle - \left\langle{f}_{2} , {f}_{2}\right\rangle + 2 \left\langlep , {f}_{2} - {f}_{1}\right\rangle - {\delta}^{2} = 2 \delta \left\lVert p - {f}_{2} \right\rVert$

Calling $c = \left\langle{f}_{1} , {f}_{1}\right\rangle - \left\langle{f}_{2} , {f}_{2}\right\rangle - {\delta}^{2}$ we have

$2 \left\langlep , {f}_{2} - {f}_{1}\right\rangle + c = 2 \delta \left\lVert p - {f}_{2} \right\rVert$

Squaring both sides

${\left(2 \left\langlep , {f}_{2} - {f}_{1}\right\rangle + c\right)}^{2} = 4 {\delta}^{2} \left\langlep - {f}_{2} , p - {f}_{2}\right\rangle$

Making the pertinent substitutions we get at

$\frac{11}{25} {x}^{2} - {y}^{2} - 11 = 0$