# How do you find the center, vertices, foci and asymptotes of 4x^2-y^2-24x-4y+28=0?

##### 1 Answer

Center $\left(h , k\right) = \left(3 , - 2\right)$
Vertex $\left(h + a , k\right) = \left(4 , - 2\right)$ and $\left(h - a , k\right) = \left(2 , - 2\right)$
Foci $\left(h + c , k\right) = \left(5.23 , - 2\right)$ and $\left(h - c , k\right) = \left(0.77 , - 2\right)$
Asymptotes $y = 2 x - 8$ and $y = - 2 x + 4$

#### Explanation:

From the given equation
$4 {x}^{2} - {y}^{2} - 24 x - 4 y + 28 = 0$

rearrange first so that the variables are together

$4 {x}^{2} - 24 x - {y}^{2} - 4 y + 28 = 0$

Perform completing the square

$4 \left({x}^{2} - 6 x\right) - \left({y}^{2} + 4 y\right) + 28 = 0$

$4 \left({x}^{2} - 6 x + 9 - 9\right) - \left({y}^{2} + 4 y + 4 - 4\right) + 28 = 0$

$4 {\left(x - 3\right)}^{2} - 36 - {\left(y + 2\right)}^{2} + 4 + 28 = 0$

$4 {\left(x - 3\right)}^{2} - {\left(y + 2\right)}^{2} - 4 = 0$

${\left(x - 3\right)}^{2} / 1 - {\left(y + 2\right)}^{2} / 4 = 1$

From the form of hyperbola

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

$a = 1$ and $b = 2$ and $c = \sqrt{5}$

Center $\left(h , k\right) = \left(3 , - 2\right)$

Vertex
$\left(h + a , k\right) = \left(3 + 1 , - 2\right) = \left(4 , - 2\right)$
$\left(h - a , k\right) = \left(3 - 1 , - 2\right) = \left(2 , - 2\right)$

Foci
$\left(h + c , k\right) = \left(3 + \sqrt{5} , - 2\right) = \left(5.23 , - 2\right)$
$\left(h - c , k\right) = \left(3 - \sqrt{5} , - 2\right) = \left(0.77 , - 2\right)$

Asymptotes

$y = \pm \frac{b}{a} \left(x - h\right) + k$

$y = + \frac{2}{1} \left(x - 3\right) + \left(- 2\right)$
$y = 2 x - 6 - 2$
$y = 2 x - 8$
and

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

$y = - 2 x + 6 - 2$

$y = - 2 x + 4$

kindly see the graph

hyperbola-colored red
asymptotes-colored blue
center-colored violet
vertices-colored orange
foci-colored black

God bless...I hope the explanation is useful.