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Completing the Square in a Hyperbola Equation

Date: 01/04/2004 at 20:37:15
From: Christina Gesar
Subject: Completing Squares in a Hyperbola Problem

I was doing a problem for hyperbolas and it said to complete the 
squares.  What does that mean and how did the solution in the book 
happen?  The question is 4(x^2  4x) - 9(y^2 + 6y) = 101 and the
answer they give is 4(x^2  4x + 4) - 9(y^2 + 6y + 9) = 36.



Date: 01/05/2004 at 09:42:37
From: Doctor Mitteldorf
Subject: Re: Completing Squares in a Hyperbola Problem

Dear Christina,

Thanks for writing.  Let's start by looking at what it means to
complete the square, and then we'll apply it to the problem you have.  

Suppose you have an equation like this

    (x - 3)^2 = 16

Since both sides of the equation are perfect squares, you can solve it
by taking the square root of each side to get

    x - 3 = 4 or -4   
        x = 4 + 3 or -4 + 3
        x = 7 or -1

That worked pretty nicely because both sides were perfect squares. 
But you may not be so lucky when you are given the equation.  They
could give it to you in a "disguised" form.  Let's multiply out the
square on the left side to get

    (x - 3)^2 = (x - 3)(x - 3) = x^2 - 3x - 3x + 9 = x^2 - 6x + 9

The original equation said that this was equal to 16, so it might have
looked like this instead 

    x^2 - 6x + 9 = 16

Even though that's the same equation we had, it looks a little harder
than it did the other way.  We could disguise it even more by
subtracting 16 from each side to get


    x^2 - 6x - 7 = 0

Again, this is still the same equation we started with at the top, but
it sure looks harder to solve this way!  It doesn't look like a
perfect square at all, and there's nothing to take the square root of.

That's where "completing the square" comes in.  It's the way to
"un-disguise" the equation, by un-doing what we just did.

There is one really important idea to use when completing the square.
In any perfect trinomial square that starts with just x^2 (not 2x^2,
for instance), like x^2 - 6x + 9, the last term (9) is the square of
half the middle term (6/2 = 3 and 3^2 = 9).

Suppose you were just given the equation 

    x^2 - 6x - 7 = 0

Our first step is to move the -7 to the other side, which leaves room
for the number we need to complete the square.  Adding 7 to each side

    x^2 - 6x     = 7

Again, notice that the middle term is 6, so half of that is 3, and 3^2
is 9.  That means we need to add 9 to 'complete the square.'  Remember
that we need to add 9 to both sides of the equation:

    x^2 - 6x + 9 = 7 + 9

Now that we have completed the square, we can rewrite the left side as
a perfect square, while at the same time adding the 7 and 9 on the
right side to get

    (x - 3)^2 = 16

And we're back to the problem we started with!  The key step was
adding the 9 to set up x^2 - 6x + 9, which is a perfect square. 
That's why this process is called "completing the square."

What would we need to complete x^2 + 14x?  Since half of 14 is 7 and
7^2 is 49, we need to add 49 to get x^2 + 14x + 49 which is the
perfect square (x + 7)^2.

Note that these cases only work when the first term is x^2.  If
there's a number in front of the x^2, we have to first factor it out.
But we'll save that for another day!

Now let's go back to your problem with the hyperbola.  You were given

    4(x^2  4x) - 9(y^2 + 6y) = 101

There are two squares to complete in this equation, x^2 - 4x and y^2 +
6y.  Both are all set with just x^2 and y^2, so we can use our 'half
the middle term squared' rule.

For x^2 - 4x, half of 4 is 2 and 2^2 is 4, so we need to add 4.  For
y^2 + 6y, half of 6 is 3 and 3^2 is 9, so we need to add 9.  Add each
of those things to the expressions in the parentheses:

    4(x^2  4x + 4) - 9(y^2 + 6y + 9) = 101

But we can't just add stuff to one side of the equation without also
doing it on the other side to preserve the balance.  So it's tempting
to just add 4 and 9 to the right side as well.  But it's trickier than
that!  Notice that the + 4 is in parenthesis, and that there is a 4
out in front that is multiplying everything in the (  ).  So if we
distributed that first part out, we'd have 4x^2 - 16x + 16.  In other
words, the net result of adding the +4 in the (  ) was really adding
16 to the left side.  We need to addd 16 to the right side to keep the
equation balanced.

Similarly, the +9 in the second (  ) is really being multiplied by the
-9 out in front.  So the net result of that is (-9)(+9) or -81.  We
balance that by adding a -81 to the right side

    4(x^2  4x + 4) - 9(y^2 + 6y + 9) = 101 + 16 - 81

We're done with the hard part.  Now we just have to write the two
parts in (  ) as the perfect squares we completed and add together all
the numbers on the right.  Doing so gives

    4(x - 2)^2 - 9(y + 3)^2 = 36

This process of completing the square is used to solve some quadratic
equations like we did at the beginning.  It's also used in conic
sections like the hyperbola, ellipse, circle or parabola because
putting their equations into a form like what we wound up with helps
us see things like the vertices and centers more easily.

That was a lot of information, and I hope it made some sense and was
helpful!  Is there another example in your book that you can try
yourself?  That would be the next step, to put what we've just learned
into practice.

Feel free to write back if you have more questions on this process,
and good luck!

- Doctor Mitteldorf, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Conic Sections/Circles
High School Conic Sections/Circles

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