# How do you find the center of the hyperbola, its focal length, and its eccentricity if a hyperbola has a vertical transverse axis of length 8 and asymptotes of y=7/2x-3 and y=-7/2x-1?

Feb 18, 2015

Center $C \left(\frac{2}{7} , - 2\right)$, focal axis $\frac{8}{7} \sqrt{53}$, eccentricity $e = \frac{\sqrt{53}}{7}$.

Since this hyperbola has a vertical transverse axis (the "$- 1$" at the second member) with center in $C \left({x}_{c} , {y}_{c}\right)$ (not the origin because the asymptotes don't pass from the origin), the equation is:

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

First of all let's find the center that is at a point that is on the two asymptotes. So let's solve the system:

$y = \frac{7}{2} x - 3$
$y = - \frac{7}{2} x - 1$

We can make a subtraction:

$y - y = \frac{7}{2} x - 3 - \left(- \frac{7}{2} x - 1\right) \Rightarrow \frac{7}{2} x - 3 + \frac{7}{2} x + 1 = 0$

$7 x = 2 \Rightarrow x = \frac{2}{7}$

and:

$y = \frac{7}{2} \cdot \frac{2}{7} - 3 \Rightarrow y = - 2$.

$C \left(\frac{2}{7} , - 2\right)$

Since the asymptotes have slope:

$m = \pm \frac{b}{a}$,

and the vertical axis is $2 b$, than:

$\frac{b}{a} = \frac{7}{2}$

$2 b = 8 \Rightarrow b = 4$

so:

$\frac{4}{a} = \frac{7}{2} \Rightarrow a = 4 \cdot \frac{2}{7} \Rightarrow a = \frac{8}{7}$.

The equation of the hyperbola is:

${\left(x - \frac{2}{7}\right)}^{2} / \left(\frac{64}{49}\right) - {\left(y + 2\right)}^{2} / 16 = - 1$.

The focal axis is named $2 c$, such as ${c}^{2} = {a}^{2} + {b}^{2}$, so:

$c = \sqrt{\frac{64}{49} + 16} = \sqrt{\frac{64 + 784}{49}} = \sqrt{\frac{848}{49}} = \sqrt{\frac{16 \cdot 53}{49}} =$

$= \frac{4}{7} \sqrt{53}$, so the focal axis is $\frac{8}{7} \sqrt{53}$.

The eccentricity is, in this case, is: $e = \frac{c}{b} = \frac{\frac{4}{7} \sqrt{53}}{4} = \frac{\sqrt{53}}{7}$