# How do I use completing the square to convert the general equation of a hyperbola to standard form?

Oct 25, 2014

I can not remember how to deal with $x y$ so I will be demonstrating completing the square without one.

Consider the equation below

$9 {x}^{2} - 16 {y}^{2} - 36 x - 96 y - 252 = 0$

The first thing that we should do is group our $x$s and $y$s

$9 {x}^{2} - 16 {y}^{2} - 36 x - 96 y - 252 = 0$
$\implies \left(9 {x}^{2} - 36 x\right) + \left(- 16 {y}^{2} - 96 y\right) - 252 = 0$

Now, let's make our work easier by factoring out ${x}^{2}$'s and ${y}^{2}$'s coefficient

$\left(9 {x}^{2} - 36 x\right) + \left(- 16 {y}^{2} - 96 y\right) - 252 = 0$

$\implies 9 \left({x}^{2} - 4 x\right) + - 16 \left({y}^{2} + 6 y\right) - 252 = 0$

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

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

when a = 1, we have

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

Now, in our equation,

$9 \left({x}^{2} - 4 x\right) + - 16 \left({y}^{2} + 6 y\right) - 252 = 0$

We want ${x}^{2} - 4 x$ and ${y}^{2} + 6 y$ to be perfect squares.
In order to do that, we need to add a 3rd element.

We know that in ${x}^{2} - 4 x$,

$2 b = - 4$
$b = - 2$

For ${x}^{2} - 4 x$ to be a perfect square, we need to add ${b}^{2} = 4$

Meanwhile, for ${y}^{2} + 6 y$,

$2 b = 6$
$b = 3$

For ${y}^{2} + 6 y$ to be a perfect square, we need to add ${b}^{2} = 9$

$9 \left({x}^{2} - 4 x\right) + - 16 \left({y}^{2} + 6 y\right) - 252 = 0$

$\implies 9 \left({x}^{2} - 4 x + 4\right) + - 16 \left({y}^{2} + 6 y + 9\right) - 252 = 0$

We added something in the left-hand side of the equation.
Since we our dealing with an equality, we need to maintain the
equality.
We can do this by adding the same value in the right-hand side of the equation or by subtracting the same value in the left-hand side.

For this demonstration, I will subtract the same value in the left-hand side

$\implies 9 \left({x}^{2} - 4 x + 4\right) + - 16 \left({y}^{2} + 6 y + 9\right) - 252 = 0$

$\implies 9 \left({x}^{2} - 4 x + 4\right) + - 16 \left({y}^{2} + 6 y + 9\right) - 252 - 9 \left(4\right) - - 16 \left(9\right) = 0$

Now that we have our perfect square trinomials and our equality remains to be one, we can proceed to simplifying the equation

$9 \left({x}^{2} - 4 x + 4\right) + - 16 \left({y}^{2} + 6 y + 9\right) - 252 - 9 \left(4\right) - - 16 \left(9\right) = 0$

$\implies 9 {\left(x - 2\right)}^{2} + - 16 {\left(y + 3\right)}^{2} - 252 - 9 \left(4\right) - - 16 \left(9\right) = 0$
$\implies 9 {\left(x - 2\right)}^{2} + - 16 {\left(y + 3\right)}^{2} - 252 - 36 - - 144 = 0$
$\implies 9 {\left(x - 2\right)}^{2} + - 16 {\left(y + 3\right)}^{2} - 144 = 0$
$\implies 9 {\left(x - 2\right)}^{2} + - 16 {\left(y + 3\right)}^{2} = 144$

$\implies \frac{9 {\left(x - 2\right)}^{2} + - 16 {\left(y + 3\right)}^{2} = 144}{144}$

$\implies \frac{9 {\left(x - 2\right)}^{2}}{144} + \frac{- 16 {\left(y + 3\right)}^{2}}{144} = \frac{144}{144}$

$\implies \frac{{\left(x - 2\right)}^{2}}{16} + \frac{- {\left(y + 3\right)}^{2}}{9} = 1$

$\implies {\left(x - 2\right)}^{2} / 16 - {\left(y + 3\right)}^{2} / 9 = 1$

Now we have transformed our general equation to standard form.
Take note that we can do the same trick to transform other conic sections into their standard form.