# How do you write the standard form of the hyperbola 9x^2-4y^2-90x+32y-163=0?

Jun 1, 2018

#### Answer:

$\frac{{\left(x - 5\right)}^{2}}{{6}^{2}} - \frac{{\left(y - 4\right)}^{2}}{{9}^{2}} = 1$

#### Explanation:

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

$9 {x}^{2} - 4 {y}^{2} - 90 x + 32 y - 163 = 0$

$\Rightarrow \left(9 {x}^{2} - 90 x\right) - \left(4 {y}^{2} - 32 y\right) = 163$

$\Rightarrow \textcolor{g r e e n}{9} \left({x}^{2} - 10 x\right) \textcolor{\lim e}{- 4} \left({y}^{2} - 8 y\right) = 163$

$\Rightarrow \textcolor{g r e e n}{9} \left({x}^{2} - 10 x \textcolor{b l u e}{+ 25}\right) - \textcolor{\lim e}{4} \left({y}^{2} - 8 y \textcolor{red}{+ 16}\right) = 163 + \textcolor{g r e e n}{9} \cdot \textcolor{b l u e}{25} + \left(\textcolor{\lim e}{- 4}\right) \cdot 16$

$\Rightarrow \textcolor{g r e e n}{9} {\left(x - 5\right)}^{2} - \textcolor{\lim e}{4} {\left(y - 4\right)}^{2} = 324$

$\Rightarrow \frac{\textcolor{g r e e n}{9} {\left(x - 5\right)}^{2}}{324} - \frac{\textcolor{\lim e}{4} \left(y - 4\right) 62}{324} = 1$

$\Rightarrow \frac{{\left(x - 5\right)}^{2}}{36} - \frac{{\left(y - 4\right)}^{2}}{81} = 1$

$\Rightarrow \frac{{\left(x - 5\right)}^{2}}{{6}^{2}} - \frac{{\left(y - 4\right)}^{2}}{{9}^{2}} = 1$