# How do you find the standard form given 4x^2+9y^2-8x+54y+49=0?

Feb 24, 2017

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

#### Explanation:

$4 {x}^{2} + 9 {y}^{2} - 8 x + 54 y + 49 = 0$

${\left(2 x\right)}^{2} + \left(3 {y}^{2}\right) - 8 x + 54 y + 49 = 0$

rearrange the equation,
$\left[{\left(2 x\right)}^{2} - 8 x\right] + \left[\left(3 {y}^{2}\right) + 54 y\right] + 49 = 0$

$\left[{\left(2 x - 2\right)}^{2} - {2}^{2}\right] + \left[\left(3 y + {9}^{2}\right) - {9}^{2}\right] + 49 = 0$

${\left(2 x - 2\right)}^{2} + \left(3 y + {9}^{2}\right) - 4 - 81 + 49 = 0$

${\left(2 x - 2\right)}^{2} + {\left(3 y + 9\right)}^{2} - 36 = 0$

${\left(2 x - 2\right)}^{2} + {\left(3 y + 9\right)}^{2} - 36 = 0$

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