General Form of the Equation
Key Questions

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 lefthand 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 righthand side of the equation or by subtracting the same value in the lefthand side.For this demonstration, I will subtract the same value in the lefthand 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. 
Let us find the slant asymptotes of a hyperbola of the form
#x^2/a^2y^2/b^2=1# .By subtracting
#x^2/a^2# ,#=>y^2/b^2=x^2/a^2+1# by multiplying by
#b^2# ,#=> y^2=b^2/a^2 x^2b^2# by taking the squareroot,
#=> y=pm sqrt{b^2/a^2 x^2b^2}# For large
#x# ,#b^2# in the squareroot is negligible,#y=pm sqrt{b^2/a^2 x^2b^2}approx pm sqrt{b^2/a^2 x^2}=pm b/a x# Hence, the slant asymptotes are
#y=pm b/a x# .
I hope that this was helpful.

Well a hyperbola will look something like this:
So in general terms we can say that an hyperbolic equation will be in the terms
or
So you can see also from this that it may be in a form like this:
#(x/a + y/b)(x/a  y/b) = 1# or
#(x/a + y/b)(x/a  y/b) = 1# 
The general form of a hyperbola is
#Ax^2 + Bx + Cy^2 + Dy + E = 0# where either A or C is negative (but never both)