# How do you find the standard form of 4x^2 - 5y^2 - 16x - 30y - 9 = 0?

Dec 17, 2015

${\left(y + 3\right)}^{2} / {2}^{2} - {\left(x - 2\right)}^{2} / {\left(\sqrt{5}\right)}^{2} = 1$

#### Explanation:

The standard form of a hyperbola is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - h\right)}^{2} / \left({a}^{2}\right) - {\left(y - k\right)}^{2} / \left({b}^{2}\right) = 1$

Given
$\textcolor{w h i t e}{\text{XXX}} 4 {x}^{2} - 5 y - 16 x - 30 y - 9 = 0$

Group the $x$ terms and the $y$ terms separately and transfer the constant to the right side of the equation.
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{4 {x}^{2} - 16 x} - \textcolor{b l u e}{5 {y}^{2} + 30 y} = \textcolor{g r e e n}{9}$

Extract the constant factors from each of the $x$ and $y$ sub-expressions.
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{4 \left({x}^{2} - 4 x\right)} - \textcolor{b l u e}{5 \left({y}^{2} + 6 y\right)} = \textcolor{g r e e n}{9}$

Complete the squares for each of the $x$ and $y$ sub-expressions
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{4 \left({x}^{2} - 4 x + {2}^{2}\right)} - \textcolor{b l u e}{5 \left({y}^{2} + 6 y + {3}^{2}\right)} = \textcolor{g r e e n}{9} + \textcolor{red}{16} - \textcolor{b l u e}{45}$

$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{4 {\left(x - 2\right)}^{2}} - \textcolor{b l u e}{5 {\left(y + 3\right)}^{2}} = - 20$

Dividing both sides by $\left(- 20\right)$
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{- {\left(x - 2\right)}^{2} / 5} \textcolor{b l u e}{+ {\left(y + 3\right)}^{2} / 4} = 1$

Convert the denominators into squares and re-order the left side terms to avoid the ugly leading minus sign.
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{\left(y + 3\right)}^{2} / {2}^{2}} - \textcolor{red}{{\left(x - 2\right)}^{2} / {\left(\sqrt{5}\right)}^{2}} = 1$

Note that the reversal of the $x$ and $y$ terms from the normal form implies that the transverse axis is vertical.
graph{4x^2-5y^2-16x-30y-9=0 [-9.14, 19.34, -9.06, 5.18]}