# How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola 4x^2-25y^2-8x-96=0?

Oct 27, 2016

The vertices are $\left(6 , 0\right)$ and $\left(- 4 , 0\right)$
The equations of the asymptotes are $y = \frac{2}{5} \left(x - 1\right)$ and $y = - \frac{2}{5} \left(x - 1\right)$
The foci are $\left(1 + \sqrt{29} , 0\right)$ and $\left(1 - \sqrt{29} , 0\right)$

#### Explanation:

The equation is $4 {x}^{2} - 8 x - 25 {y}^{2} = 96$

$4 \left({x}^{2} - 2 x\right) - 25 {y}^{2} = 96$

$4 \left({x}^{2} - 2 x + 1\right) - 25 {y}^{2} = 96 + 4$

$4 {\left(x - 1\right)}^{2} - 25 {y}^{2} = 100$
Dividing by 100

${\left(x - 1\right)}^{2} / 25 - {y}^{2} / 4 = 1$

${\left(x - 1\right)}^{2} / {5}^{2} - {y}^{2} / {2}^{2} = 1$
This is the standard equation of a left right hyperbola
${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$

so, the center is $\left(h , k\right) = \left(1 , 0\right)$
the vertices ae $\left(h + a , k\right) = \left(1 + 5 , 0\right) = \left(6 , 0\right)$
and $\left(h - a , k\right) = \left(1 - 5 , 0\right) = \left(- 4 , 0\right)$
The slopes of the asymptotes are $\pm \frac{b}{a} = \pm \frac{2}{5}$
And the equations of the asymptotes are $y = \pm \frac{b}{a} \left(x - h\right)$
$\implies$$y = \pm \frac{2}{5} \left(x - 1\right)$
To determine the foci, we need $c = \pm \sqrt{{a}^{2} + {b}^{2}} = \pm \sqrt{25 + 4} = \pm \sqrt{29}$
So the foci are $\left(h + c , k\right) = \left(1 + \sqrt{29} , 0\right)$
and $\left(h - c , k\right) = \left(1 - \sqrt{29} , 0\right)$