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

1 Answer
Oct 25, 2014

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 xs and ys

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.