# How do you find the standard form of 3x^2+5y^2-12x+80y+40=0?

Jan 14, 2017

$\frac{{\left(x - 2\right)}^{2}}{5} + \frac{{\left(y + 8\right)}^{2}}{3} = 24.8$

#### Explanation:

$3 {x}^{2} + 5 {y}^{2} - 12 x + 80 y + 40 = 0$

Combine x terms and y terms together:
$\left[3 {x}^{2} - 12 x\right] + \left[5 {y}^{2} + 80 y\right] + 40 = 0$

Complete the square for x terms and y terms separately:
$3 \left[{x}^{2} - 4 x\right] + 5 \left[{y}^{2} + 16 y\right] + 40 = 0$

$3 \left[{\left(x - 2\right)}^{2} - 4\right] + 5 \left[{\left(y + 8\right)}^{2} - 64\right] - 40 = 0$

$3 {\left(x - 2\right)}^{2} - 12 + 5 {\left(y + 8\right)}^{2} - 320 - 40 = 0$

Combine constants and simplify:
$3 {\left(x - 2\right)}^{2} + 5 {\left(y + 8\right)}^{2} - 372 = 0$

$\textcolor{red}{\frac{1}{15} \cdot} \left(3 {\left(x - 2\right)}^{2} + 5 {\left(y + 8\right)}^{2}\right) = \left(372\right) \textcolor{red}{\cdot \frac{1}{15}}$

$\frac{3 {\left(x - 2\right)}^{2}}{15} + \frac{5 {\left(y + 8\right)}^{2}}{15} = \frac{124}{5}$

$\frac{{\left(x - 2\right)}^{2}}{5} + \frac{{\left(y + 8\right)}^{2}}{3} = 24.8$