# How do you write a standard form equation for the hyperbola with y^2 - 4y - 4x^2 + 8x = 4?

Jul 16, 2018

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

#### Explanation:

The general equation for a hyperbola is

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

or

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

Here, we have

${y}^{2} - 4 y - 4 {x}^{2} + 8 x = 4$

Complete the square

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

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

Dividing by $4$

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

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

The center is $C = \left(1 , 2\right)$ and the vertices are $A = \left(1 , 0\right)$ and $A ' = \left(1 , 4\right)$

graph{((y-2)^2/4-(x-1)^2/1-1)=0 [-20.43, 20.11, -9, 11.28]}

Jul 16, 2018

Standard form of vertical hyperbola equation is
${\left(y - 2\right)}^{2} / {2}^{2} - {\left(x - 1\right)}^{2} / \left({1}^{2}\right) = 1$

#### Explanation:

${y}^{2} - 4 y - 4 {x}^{2} + 8 x = 4$ or

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

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

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

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

The standard form of equation of vertical hyperbola is

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

Therefore, the standard form of vertical hyperbola equation

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

graph{y^2-4 y -4 x^2+ 8 x -4=0 [-40, 40, -20, 20]} [Ans]