All the terms are of the first degree (no squares, square roots, cubes,etc) yet when graphed, it looks as if it is an hyperbola. However, equations for the general form of hyperbolas are of the second degree (as well as all the other conic sections).
I believe this would be called a 2 variable equation but it is a special case. For example, X + Y =24 is also a 2 varaible equation but when graphed it is linear.
It is, in fact, a hyperbola. You can think of it as degree two because the monomial xy has degree 2 (degree of x + degree of y = 1+1 = 2). This hyperbola is rotated 45[sup]o[/sup] from the usual horizontal/vertical orientation you may be used to. If you transform it by rotating it 45[sup]o[/sup] (in either direction), you’ll get a hyperbola equation of the form you’re more used to seeing.
As Cabbage says, it is a degree-two equation. After rearranging it to XY - 24 = 0, I’d go further and call it a quadratic bivariate polynomial in X and Y, but that’s just me.
There’s more than you wished to know about polynomial taxonomy here.
It’s just y = x^-1 with a scaling transformation, so it should be apparent it’s a hyperbola.
My last post was a bit rushed; now that a have a bit more time, I thought it might help to go into a bit more detail about the transformation.
We want to take the hyperbola xy=24 and rotate it 45[sup]o[/sup], counterclockwise (ccw), let’s say. Rotation is a linear transformation, so I only need to consider what the rotation does to the unit basis vectors (1,0) and (0,1). I’ll use x, y variables for points in the domain of the transformation, and s, t for the range.
If I rotate (1,0) 45[sup]o[/sup] ccw, it moves to the point (1/sqrt(2), 1/sqrt(2)).
If I rotate (0,1) 45[sup]o[/sup] ccw, it moves to (-1/sqrt(2), 1/sqrt(2)).
So an arbitrary point (x,y) gets moved to (s,t) = (x/sqrt(2) - y/sqrt(2), x/sqrt(2) + y/sqrt(2)). I need to find x and y in terms of s and t, in order to know the appropriate substituion. I get:
(x,y) = (s/sqrt(2) + t/sqrt(2), -s/sqrt(2) + t/sqrt(2)). Make this substitution into our hyperbola:
xy = 24
becomes
(s/sqrt(2) + t/sqrt(2))(-s/sqrt(2) + t/sqrt(2)) = 24
-s[sup]2[/sup]/2 + t[sup]2[/sup]/2 = 24
t[sup]2[/sup]/48 - s[sup]2[/sup]/48 = 1
which is probably the standard form you’re used to seeing.
If it’s anything, it’s of degree two. Since all the coefficients are integers, it can fit into Z[X,Y], which is a filtered Z-algebra. That is, it’s a Z-module (Abelian group) M which is the projective limit of a sequence of Z-modules M[sub]n[/sub] and equipped with a Z-bilinear map from M x M to M such that M[sub]i[/sub] x M[sub]j[/sub] falls within M[sub]i+j[/sub]. In this case, the usual (and I believe essentially unique) filtration has M[sub]n[/sub] as the Z-linear span of monomials in X and Y such that the sum of the exponents on X and Y is less than or equal to n. The polynomial XY-24 thus has filtration degree 2.
Now, strictly speaking Z[X,Y] is also a graded algebra, which is pretty much the same except that it’s expressed as the direct sum of a number of Z-modules A[sub]n[/sub] rather than as the projective limit and A[sub]i[/sub] x A[sub]j[/sub] maps exactly into A[sub]i+j[/sub]. The standard grading has A[sub]n[/sub] as the linear span of all monomials in X and Y such that the sum of the exponents is exactly n. The polynomial XY-24 has no degree under this view since it doesn’t fit into any one of these A[sub]n[/sub]. Any graded algebra, however, has a filtration “by maximal degree”, which recovers the above filtration. XY-24 has “maximal degree” 2.
There’s also a bigrading on Z[X,Y], but there is no canonical “maximal degree” filtration on a multigraded module, so that’s no use.
Bytegeist
So then it could be considered a 2 variable quadratic equation?
or as you said a quadratic bivariate polynomial.
It is funny there is not more information about this (on the 'Net or in books).
Such an equation crops up in Algebra I. (which I took a LONG time ago).
Example:
2 numbers when multiplied equal 24 and when added equal 10. What are they?
-
X * Y =24
-
X + Y =10
-
Y = 24 / X and substituting this into equation 2 we get:
X + 24/X =10
X^2 + 24 = 10X
X^2 -10X + 24 = 0
X=6 X=4
Sorry for posting such a trivial example, but this is the kind of problem we used to do in 9th grade algebra.
SO, as I said previously, it is funny that such an equation is simple enough for 9th grade algebra and yet it is rather adavnced too.
Also, it is strange there is so little information about it.
The standard hyperbola that’s taught in high-school algebra is a special case where the line between the foci is parallel to one of the axes. If you consider a more general case, you’ll get an equation of the form ax[sup]2[/sup] + bxy + cy[sup]2[/sup] + dx + ey + f = 0.
And of course, the hyperbola given by xy - 24 = 0 has foci on the line y = x which are equidistant from the origin.
Please humor me …
Would x[sup]3/2[/sup] y[sup]1/2[/sup] = K also be a hyperbola? Rotated 22.5 degrees from the vertical?
KarlGauss, no, that wouldn’t be a hyperbola. x[sup]3/2[/sup]y[sup]1/2[/sup] isn’t a polynomial in x and y, first of all; the exponents in a polynomial must be nonegative integers, not rational numbers.
It’d probably help to see that it’s not a hyperbola by playing around with it a bit:
x[sup]3/2[/sup]y[sup]1/2[/sup] = K
(We’ll say K is a positive real number, since it’s uninteresting otherwise).
Square both sides:
x[sup]3[/sup]y = K[sup]2[/sup]
If I include the restriction that x and y are both positive, then these two equations are equivalent (otherwise, without the restriction, the latter equation would have extra stuff in the third quadrant, where x and y are both negative). In this sense, we could consider the original equation to be fourth degree rather than second degree. The bottom line is that in this equation y varies inversely with x[sup]3[/sup] (instead of varying inversely with x, like the hyperbola in the OP), so it’s not a hyperbola. I don’t know if there’s a standard name for this family of curves.
Ah, indeed, a fourth degree polynomial. [sub][sup]I probably once knew that. Really.[/sup][/sub]
Thank you for your patience and teaching.
No, x[sup]3/2[/sup] y[sup]1/2[/sup] - K is an element of maximal degree 4 in R[x[sup]1/2[/sup],y[sup]1/2[/sup]], where R is a ring containing K.
Should we tell them about what means what in the quadratic terms and such? I think so.
Break the general quadratic in two variables into the quadratic part (ax[sup]2[/sup] + bxy + cy[sup]2[/sup]), the linear part (dx+ey), and the constant part (f).
If we write the pair of x and y as a column vector, the linear part (one should remember) is given by the matrix product with the pair of d and e as row vectors.
(x)
(d e) ( ) = (dx+ey)
(y)
Similarly, the quadratic part can be expressed in matrix notation.
( a b/2) (x)
(x y) ( ) ( ) = (ax[sup]2[/sup] + bxy + cy[sup]2[/sup])
(b/2 c ) (y)
Now, the thing that determines how the graph behaves is the determinant if this 2x2 matrix: ac-b[sup]2[/sup]/4. If positive, the graph is an ellipse. If zero, the graph is a parabola (consider when b and c are both zero). If negative, the graph is a hyperbola. Work this out yourself for XY-24 and X[sup]2[/sup]-Y[sup]2[/sup]-24.
Now explore what happens when you use three variables.