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Approaching Asymptotes of Hyperbolas

Date: 04/24/2011 at 08:10:04
From: Donald
Subject: Derivations of standard hyperbolas asymptotes

How do you derive the equations for the asymptotes of the standard
hyperbolas?

   y = +/-(b/a)x
   y = +/-(a/b)x

Solving for y, I got it down to:

   y = +/-(b/a) sqrt(x^2 - a^2)

Then, letting x go to infinity, the a^2 is rendered insignificant, so 

   y = +/-(b/a) sqrt((x^2))
   
This gives

   y = +/-(b/a)x

But, how does this prove that y = +/- (b/a)x is an asymptote(s)? I need
clarity here.

I have been to site after site, and looked in books, and I still can't
find an explanation. They all just state it and how to use it, but offer
no proof. Could you help me with this?



Date: 04/24/2011 at 22:46:44
From: Doctor Peterson
Subject: Re: Derivations of standard hyperbolas asymptotes

Hi, Donald.

What you've done is a good informal demonstration of the idea; we can make
it a little more convincing by saying it this way:

   y = +/-b/a sqrt(x^2 - a^2)
     = +/-b/a sqrt(x^2(1 - a^2/x^2))

When x is much larger than a, a^2/x^2 is much less than 1, so this will be
very close to

   y = +/-b/a sqrt(x^2) = +/-(b/a)x

For a real proof, we have to start with the definition of "asymptote";
without that, no proof is possible, since we wouldn't know what we were
trying to prove!

An asymptote is a line that is approached more and more nearly by the
curve as x increases. That is, if we have a curve y = f(x) and a line 
y = mx + b, the latter is an asymptote of the former if

     lim[x->oo](f(x) - (mx + b)) = 0

To show that y = bx/a is an asymptote of y = b/a sqrt(x^2 - a^2), we want
to show that

     lim[x->oo](b/a sqrt(x^2 - a^2) - bx/a) = 0

Is it?

     lim[x->oo](b/a sqrt(x^2 - a^2) - bx/a)
   
   = lim[x->oo](b/a (sqrt(x^2 - a^2) - x))

   = b/a lim[x->oo](sqrt(x^2 - a^2) - x)

                    (sqrt(x^2 - a^2) - x)(sqrt(x^2 - a^2) + x)
   = b/a lim[x->oo]--------------------------------------------
                                         (sqrt(x^2 - a^2) + x)

                    (x^2 - a^2)^2 - x^2
   = b/a lim[x->oo]---------------------
                    sqrt(x^2 - a^2) + x

                     x((1 - a^2/x^2)^2 - 1)
   = b/a lim[x->oo]--------------------------
                    x(sqrt(1 - a^2/x^2) + 1)

                     (1 - a^2/x^2)^2 - 1
   = b/a lim[x->oo]-----------------------
                    sqrt(1 - a^2/x^2) + 1

                     (1)^2 - 1
   = b/a lim[x->oo]-------------
                    sqrt(1) + 1

      0
   = ---
      2   

   = 0

So the curve does in fact approach the line.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 04/26/2011 at 06:34:47
From: Donald
Subject: Thank you (Derivations of standard hyperbolas asymptotes)

I understand, and thank-you so much for the precision and the clarity.

Blessings to you and yours,

D.
Associated Topics:
High School Basic Algebra

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