# How do you write 25y^2 + 9x^2 - 50y - 54x = 119 in standard form?

May 3, 2015

$25 {y}^{2} + 9 {x}^{2} - 50 y - 54 x = 119$

Consider the tow sub-expressions from the left side of this equation:

1. Terms involving $\textcolor{red}{y}$
$\textcolor{red}{25 {y}^{2} - 50 y}$
$\textcolor{red}{= 25 \left({y}^{2} - 2 y\right)}$
$\textcolor{red}{= 25 \left({y}^{2} - 2 y + 1\right) - 25}$
$\textcolor{red}{= {5}^{2} {\left(y - 1\right)}^{2} - 25}$

2. Terms involving $\textcolor{b l u e}{x}$
$\textcolor{b l u e}{9 {x}^{2} - 54 x}$
$\textcolor{b l u e}{= 9 \left({x}^{2} - 6 x + {\left(- 3\right)}^{2}\right) - 81}$
$\textcolor{b l u e}{= {3}^{2} {\left(x - 3\right)}^{2} - 81}$

$25 {y}^{2} + 9 {x}^{2} - 50 y - 54 x = 119$
$25 {y}^{2} - 50 y + 9 {x}^{2} - 54 x = 119$
$= \textcolor{red}{{5}^{2} {\left(y - 1\right)}^{2} - 25} + \textcolor{b l u e}{{3}^{2} {\left(x - 3\right)}^{2} - 81} = 119$
$= {5}^{2} {\left(y - 1\right)}^{2} + {3}^{2} {\left(x - 3\right)}^{2} = 225$
or
$= {\left(25 y - 25\right)}^{2} + {\left(9 x - 27\right)}^{2} = {15}^{2}$
or
some variant of this depending upon your local definition of "standard form" for an ellipse.