Approaching Asymptotes of HyperbolasDate: 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. |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/