# How do I use completing the square to rewrite the equation of an ellipse in standard form?

Oct 19, 2014

To demonstrate, let's try transforming the equation below into standard form.

$4 {x}^{2} + 9 {y}^{2} + 72 y - 24 x + 144 = 0$

The first thing to do is group $x$s and $y$s together

$\left(4 {x}^{2} - 24 x\right) + \left(9 {y}^{2} + 72 y\right) + 144$

Factor out ${x}^{2}$'s and ${y}^{2}$'s coefficient

$4 \left({x}^{2} - 6 x\right) + 9 \left({y}^{2} + 8 y\right) + 144 = 0$

Before we continue, let's recall what happens when a binomial is squared

${\left(a x + b\right)}^{2} = {a}^{2} {x}^{2} + 2 a b x + {b}^{2}$

For our problem, we want ${x}^{2} - 6 x$ and ${y}^{2} + 8 y$ to be perfect squares

For $x$, we know that

${a}^{2} = 1$
$\implies a = 1 , a = - 1$

$2 a b = - 6$
$2 \left(1\right) b = - 6$
$2 b = - 6$
$b = - 3$
$\implies {b}^{2} = 9$

Note that substituting $a = - 1$ will result to the same ${b}^{2}$

Hence, to complete the square, we need to add $9$

Meanwhile, for $y$

${a}^{2} = 1$
$\implies a = 1 , a = - 1$

$2 a b = 8$
$2 \left(1\right) b = 8$
$2 b = 8$
$b = 4$
$\implies b = 16$

Now, let's add our ${b}^{2}$s into the equation.

$4 \left({x}^{2} - 6 x\right) + 9 \left({y}^{2} + 8 y\right) + 144 = 0$

$4 \left({x}^{2} - 6 x + 9\right) + 9 \left({y}^{2} + 8 y + 16\right) + 144 = 0$

We added something into the left-hand side, to retain the "equality", we need to add the same value to the right-hand side (or substract the value again from the left-hand side)

$4 \left({x}^{2} - 6 x + 9\right) + 9 \left({y}^{2} + 8 y + 16\right) + 144 = 0 + 4 \left(9\right) + 9 \left(16\right)$
$4 \left({x}^{2} - 6 x + 9\right) + 9 \left({y}^{2} + 8 y + 16\right) + 144 = 180$

$\implies 4 {\left(x - 3\right)}^{2} + 9 {\left(y + 4\right)}^{2} + 144 = 180$
$\implies 4 {\left(x - 3\right)}^{2} + 9 {\left(y + 4\right)}^{2} = 180 - 144$
$\implies 4 {\left(x - 3\right)}^{2} + 9 {\left(y + 4\right)}^{2} = 36$

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

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