Talk:Hyperbola

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[edit] Old comments

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)

[edit] 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, (x-h)(y-k) = c \,, 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)

[edit] Hyperbolic functions

I removed the reference to hyperbolic functions parametrizing hyperbolas. The most obvious way that is done is only valid in special cases. It can be done more generally, but any hyperbola can be parametrized by rational functions. In fact, any conic can; these are curves of genus zero and are therefore rationally parametrizable. The hyperbolic functions are analogous to the circular functions, and that could be further explained, I suppose. Gene Ward Smith 05:28, 20 July 2005 (UTC)

[edit] "Exaggeration"?

The American Heritage Dictionary translates the Greek as "a throwing beyond, excess (from the relationship between the line joining the vertices of a conic and the line through its focus and parallel to its directrix)". This provides more insight. Compare to parabola, "comparison, application, parabola (from the relationship between the line joining the vertices of a conic and the line through its focus and parallel to its directrix)", and ellipse, "to fall short (from the relationship between the line joining the vertices of a conic and the line through the focus and parallel to the directrix of a conic)". All three definitions implicitly refer to the eccentricity of the conic. KSmrq 03:28, 2005 August 11 (UTC)

Isn't eccentricity simply e=c/a ? Also, I see no mention of vertices or latus rectum, I thought that was an element of a hyperbola... --69.113.7.18

[edit] The simplest case

Shouldn't the simplest and most common case of a hyperbola be mentioned and shown, I mean the function f(x) = 1/x?

nl:Gebruiker:Sokpopje

Done. Geekdog 13:50, 29 January 2007 (UTC)

IN FACT, I SEE THIS MENTIONED, BUT I DO NOT SEE THIS SHOWN. AND I AGREE THAT THIS SHOULD BE DONE Athkalani 20:41, 27 May 2007 (UTC)

Can you include the proof of the relationship between a, b, and c in a hyperbola is c^2=a^2+b^2 ? 68.83.181.181 03:10, 8 January 2007 (UTC)

[edit] Ambigenal hyperbola

Does anyone actually have a clue what the definition given means? I'm not claiming to be the smartest person in the world, but I'm doing a PhD in maths, and if I can't make head or tail of that description then I suspect other people may not be able to either. A picture would help no end. All the other results for "Ambigenal hyperbola" on Google give the same definition, so I have been unable to find any clarification. As a side note, why are all the diagrams of such simple hyperbolas? They all look symmetric, which surely isn't true in the general case. Geekdog 16:05, 28 January 2007 (UTC)

I removed the "definition". It was quite unclear, and the term isn't used elsewhere in the article, so it wasn't needed. On your side note, all hyperbolas are symmetric, just not necessarily with respect to the axes, but linear transformation can make them so. So, geometrically it's all the same. A graph of y=1/x wouldn't be bad to include for an example of a "rotated" hyperbola. Cheers, Doctormatt 18:00, 28 January 2007 (UTC)
Okay, thanks for that. I'm still confused though (admittedly, geometry is not one of strong points). I've made a couple of very rough diagrams that try to explain what the problem I'm having is. This image attempts to illustrate a case where the hyperbola generated is obviously symmetric. Here is a similar case where the hyperbola is not obviously symmetric. It is not necessarily rotated (depending on coordinate choice, of course). When considered on the plane that's being used to generate the hyperbola, the two sides of the hyperbola do not seem (to my mind) to have the same curvature. When projected onto a plane containing the axis of the cone the two sides of the hyperbola are not equidistant from the centre point of the cone. I can't intuit whether they have the same curvature but I'm guessing not - of course, if they do have the same curvature then a simple translation will make the hyperbola symmetric.
You are saying that a linear transformation of this second case can produce a coordinate system where both sides of the hyperbola are symmetric, right? If so, this is not at all obvious to me.
Apologies for the low quality sketches, I hope they're enough to get my point across. Geekdog 19:56, 28 January 2007 (UTC)
I see what you mean: with your sketches, it is hard to see the symmetry. Hyperbolas are always symmetric, even the "non-vertical plane through the cone ones". Try looking at the illustrations on conic section; that might help. (It would be great to have a short animation showing the plane through the cone, and then moving so that the plane is viewed perpendicularly so you can see the symmetry). Or, consider the definition of the conic section as point loci; hyperbolas are the locus of points with the property that the absolute difference of the distances from each point to two fixed points (foci) is constant. This definition makes the symmetry most clear to me. Cheers, Doctormatt 22:03, 28 January 2007 (UTC)
Thanks, the sketch on the conic sections page made things clearer. I added a graph of y=1/x. Geekdog 13:51, 29 January 2007 (UTC)

[edit] The parametric equations are very wrong

it is written beneath the parametric equations that a is the semimajor axis and b is semiminor. This is true for a and b in the cartesian equations, but is not true for the parametric equations given. In the parametric equations, each instance of 'a' should be replaced with 'b' and the other way round to be consistent with the statement beneath. —The preceding unsigned comment was added by 144.136.38.19 (talk) 05:56, 21 March 2007 (UTC).

[edit] Pluralisation

English uses many irregular plural forms, a greek root is a common cause; the germanic pluralisation of greek words has become more common, which does not mean it is correct. By derivation, the plurals of hyperbola, parabola and formula should be hyperbolae, parabolae and formulae and, as the page on english pluralisation remarks, "correctly formed Latin plurals are the most acceptable, and indeed are often required, in academic and scientific contexts.". Diysurgery 02:44, 1 October 2007 (UTC)

I checked Merriam-Webster online, and it indicates that both are acceptable. This claim at english pluralisation is uncited; it is quite possibly just one person's opinion. I note that Mathworld suggests hyperbolas is correct ([1]) and even gives a citation (although I imagine that may be simply citing that author's usage). My trusty and worn Webster's New Collegiate Dictionary, copyright 1949, indicates hyperbolas as the only correct pluralization. This article at encarta [2] gives both. I can't find any source, besides Wikipedia, that suggests that -ae is required for hyperbola. Doctormatt 01:30, 1 October 2007 (UTC)
I've decided to focus on "formula", since it is a far more common word and the issue is (and my relevant findings are) identical. The Online Etymology Dictionary led me to the Charles Kingsley quote "Men who try to speak what they believe, are naked men fighting men quilted sevenfold in formulae.", dated 1861.
The 1982 concise OED lists both, but only uses -ae, whereas the 1996 compact OED lists both and uses -as in the entry and -ae in the entry for formulary (the english form of a latin word with a greek root, just imagine the carnage) - demonstrating the fact that, in matters of etymological minutiae, general purpose reference books simply cannot be relied upon (though I'd bet you believe me when I tell you the plural of minutia is minutiae, which neither dictionary disputes, even the laughably fallible dictionary.com agrees). Diysurgery 02:44, 1 October 2007 (UTC)

[edit] a and b used before definition

In the `Definitions' section a and b are used before they are defined later in the `Equations' section. Also the formulae they are used in are not general enough.

--Ts4079 (talk) 16:48, 28 November 2007 (UTC)

[edit] Degenerate hyperbola?

The 'degenerate hyperbola' is worth mentioning in this article. WinterSpw (talk) 05:48, 29 May 2008 (UTC)