# How do you write the hyperbola y^2-x^2/36=4 in standard form?

##### 1 Answer
Dec 12, 2016

color(green)((y-0)^2/(2^2)-(x-0)^2/(12^2)=1

#### Explanation:

The standard form for a hyperbola is
either
$\textcolor{w h i t e}{\text{XXX}} {\left(x - h\right)}^{2} / \left({a}^{2}\right) - {\left(y - k\right)}^{2} / \left({b}^{2}\right) = 1$ for a hyperbola opening left-right
or
$\textcolor{w h i t e}{\text{XXX}} {\left(y - k\right)}^{2} / \left({b}^{2}\right) - {\left(x - a\right)}^{2} / \left({a}^{2}\right) = 1$ for a hyperbola opening up-down.
In both cases the center is at $\left(h , k\right)$

Given
$\textcolor{w h i t e}{\text{XXX}} {y}^{2} - {x}^{2} / 36 = 4$
we obviously will need to transform this into the second (up-down opening) form.

$\textcolor{w h i t e}{\text{XXX}} {y}^{2} / 4 - {x}^{2} / \left(4 \cdot 36\right) = 1$

$\textcolor{w h i t e}{\text{XXX}} {y}^{2} / {2}^{2} - {x}^{2} / \left({12}^{2}\right) = 1$

or (if you wish to make the center explicit)
$\textcolor{w h i t e}{\text{XXX}} {\left(y - 0\right)}^{2} / \left({2}^{2}\right) - {\left(x - 0\right)}^{2} / \left({12}^{2}\right) = 1$

While, not technically in "standard form", this could be simplified as:
$\textcolor{w h i t e}{\text{XXX}} {y}^{2} / 4 - {x}^{2} / 144 = 1$