# How do you write the standard form of the hyperbola -x^2+y^2-18x-14y-132=0?

Oct 13, 2017

Standard form of the hyperbola equation :
${\left(y - 7\right)}^{2} / 100 - {\left(x + 9\right)}^{2} / 100 = 1$.

#### Explanation:

The standard form of a hyperbola with centre, $\left(\alpha , \beta\right)$ is $\rightarrow$

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

Therefore, the given expression can be arranged as $\rightarrow$

$- \left({x}^{2} + 18 x + 81\right) + \left({y}^{2} - 14 y + 49\right) = \left(132 - 81 + 49\right)$

$\therefore \left(x + {9}^{2}\right) - {\left(y - 7\right)}^{2} = - 100$

$\therefore \left({\left(x + 9\right)}^{2} / - 100\right) - \left({\left(y - 7\right)}^{2} / - 100\right) = 1$

$\therefore {\left(y - 7\right)}^{2} / 100 - {\left(x + 9\right)}^{2} / 100 = 1$. (Answer).

Hope it Helps:)