Talk:Hyperbola

Is this image used by permission? --Brion

It ain't local either. Aren't there laws against this type image use? --mav

IANAL, but I have the impression that case law would not be on our side if we're not using it with permission. If we got permission long long ago, in the before time, then it should be acceptable... but there's nothing in the talk page, and the pre-January edit history comments are still missing. However even if it's by permission, it doesn't look that great and we should use a local image anyway. Maybe I'll break out the old POV-Ray and make some new ones. :) --Brion

---

Why are there four different cartesian equations? How do they differ, which one should I choose, what are a, b and c? Equations without explanations are worse than useless. AxelBoldt 05:56 Jan 4, 2003 (UTC)

In the first set of two, the main difference is which way the hyperbola opens up (which direction the transverse axis is). The first one (the x term being positive) opens up horizontally, like in the picture, and the other one (with the y term being positive) opens up vertically. In these equations, a, b, h, and k are all constants.

As for the other equations, I had not seen them before, partly because they are not equations for hyperbolas, but rather equations dealing with hyperbolas, and I have not studied conics very deeply. According to my text book, c is the distance between one focus and the center. The text I have lists b² = c² - a², which I cannot seem to make equal to the formula given. The eccentricity is just what is stated in the link. Loggie 00:55, May 6, 2005 (UTC)

---

I assigned my class homework to add information to this page and show me what they added. I hope it doesn't lead to a bunch of vandalism, but I fear it may. Sorry in advance. I hope it turns out that my experiment is good for the wiki. Tjdw 21:21, 30 Apr 2004 (UTC)

Followup: It seems it was a terrible experiment. Only 5 out of 30 kids even did the homework, and of those, only 2 wrote anything useful or correct. One student plagiarized, one vandalized the page, and one wrote an incorrect process in the External Links section for isolating y on one side of the standard form equation. What do you guys think? Leave some messages on my talk page. Tjdw 22:33, 3 May 2004 (UTC)

Whoops! Make that 2 students who plagiarized. That leaves 1 who wrote correct and original content. Tjdw 22:37, 3 May 2004 (UTC)

Is this right?

A special case of the hyperbola is the equilateral or rectangular hyperbola, in which the asymptotes intersect at right angles. The rectangular hyperbola with the coordinate axes as its asymptotes is given by the equation xy=c, where c is a constant.

According to the formula given, <math>(x-h)(y-k) = c \,<math>, wouldn't xy=c anytime the center of the hyperbola is at (0, 0), not necessarily when the asymptotes intersect each other at right angles? The intercepts of the asymptotes are defined by the center (h and k), but the actual slope is defined by a and b. Loggie 01:02, May 6, 2005 (UTC)

It is right, but I corrected the other thing.--Patrick 14:27, 6 May 2005 (UTC)

I'm still slightly confused- When you say that "coordinate axes parallel to their asymptotes", are we refering to the x axis and y axis? And aren't the asymptotes of a hyperbola slanted? Can the axes of a hyperbola ever be parallel to the x and y axis? I wouldn't think so- as you made the slope of the asymptotes closer and closer to zero the hyperbola would just keep getting shallower and shallower, or steeper. I don't believe the slope of the asymptotes ever could be zero. I wish they would go into more detail when teaching me these things at school. Loggie 21:55, May 6, 2005 (UTC)

Nevermind! I got it all sorted out. Loggie 02:15, May 9, 2005 (UTC)