# How do you find the standard form given 9x^2-4y^2-36x-24y-36=0?

May 3, 2018

${\left(x - 2\right)}^{2} / {2}^{2} - {\left(y + 3\right)}^{2} / {3}^{2} = 1$ , equation of a hyperbola .

#### Explanation:

$9 {x}^{2} - 4 {y}^{2} - 36 x - 24 y - 36 = 0$ or

$9 \left({x}^{2} - 4 x\right) - 4 \left({y}^{2} + 6 y\right) = 36$ or

$9 \left({x}^{2} - 4 x + 4\right) - 4 \left({y}^{2} + 6 y + 9\right) = 36$ or

$9 {\left(x - 2\right)}^{2} - 4 {\left(y + 3\right)}^{2} = 36$ or

$\frac{9 {\left(x - 2\right)}^{2}}{36} - \frac{4 {\left(y + 3\right)}^{2}}{36} = 1$ or

${\left(x - 2\right)}^{2} / 4 - {\left(y + 3\right)}^{2} / 9 = 1$ or

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

This is standard form of the equation of a hyperbola with center

at $\left(2 , - 3\right)$

graph{9 x^2-4 y^2-36 x-24 y=36 [-80, 80, -40, 40]} [Ans]