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

I can not remember how to deal with #xy# so I will be demonstrating completing the square without one.

Consider the equation below

#9x^2 -16y^2 -36x - 96y - 252 = 0#

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

#9x^2 -16y^2 -36x - 96y - 252 = 0#
#=> (9x^2 -36x) + (-16y^2 - 96y) - 252 = 0#

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

#(9x^2 -36x) + (-16y^2 - 96y) - 252 = 0#

#=> 9(x^2 - 4x) + -16(y^2 + 6y) - 252 = 0#

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

#(ax + b)^2 = a^2x^2 + 2ab + b^2#

when a = 1, we have

#(x + b)^2 = x^2 + 2b + b^2#

Now, in our equation,

#9(x^2 - 4x) + -16(y^2 + 6y) - 252 = 0#

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

We know that in #x^2 - 4x#,

#2b = -4 #
#b = -2#

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

Meanwhile, for #y^2 + 6y#,

#2b = 6#
#b = 3#

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

#9(x^2 - 4x) + -16(y^2 + 6y) - 252 = 0#

#=> 9(x^2 - 4x + 4) + -16(y^2 + 6y + 9) - 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

#=> 9(x^2 - 4x + 4) + -16(y^2 + 6y + 9) - 252 = 0#

#=> 9(x^2 - 4x + 4) + -16(y^2 + 6y + 9) - 252 - 9(4) - -16(9) = 0#

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

#9(x^2 - 4x + 4) + -16(y^2 + 6y + 9) - 252 - 9(4) - -16(9) = 0#

#=> 9(x - 2)^2 + -16(y + 3)^2 - 252 - 9(4) - -16(9) = 0#
#=> 9(x - 2)^2 + -16(y + 3)^2 - 252 - 36 - -144 = 0#
#=> 9(x - 2)^2 + -16(y + 3)^2 - 144 = 0#
#=> 9(x - 2)^2 + -16(y + 3)^2 = 144#

#=> (9(x - 2)^2 + -16(y + 3)^2 = 144)/144#

#=> (9(x - 2)^2)/144 + (-16(y + 3)^2)/144 = 144/144#

#=> ((x - 2)^2)/16 + (-(y + 3)^2)/9 = 1#

#=> (x - 2)^2/16 - (y + 3)^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.