Since my last thread , I’ve done quite a bit of research into hyperbolae and conic sections. Not being a “math person,” some of it is over my head, but I remember enough algebra and trig to piece my way through. Rather than throw another list of questions out there, I’m trying to show the specific steps I’m taking to render a given hyperbola in standard form, hoping some of the finer math minds here on the board will be able to help me where I’m going wrong and provide any insights.
Specifically, I’m dealing with hyperbolae that fit the following form:
(1): y = (a + x[sup]2[/sup])/(b - x)
…where a and b are both positive integers. Again, for the purposes of this thread, the only hyperbolae I’m interested in fit this form.
Most of the references I have deal with hyperbolae in the following forms:
(2): Ax[sup]2[/sup] + Bxy + Cy[sup]2[/sup] + Dx + Ey + F = 0
…where B[sup]2[/sup] > 4AC
And:
(3): (y[sup]2[/sup]/a’ [sup]2[/sup]) - (x[sup]2[/sup]/b’ [sup]2[/sup]) = 1
…when the foci are on the y axis, or parametrically:
x = a’ * sec(t)
y = b’ * tan(t)
So, the goal here is to express (1) as (3) in terms of the original variables a and b
Step 1, in which we try to put (1) into standard form (2):
(a + x[sup]2[/sup])/(b - x) = y
a + x[sup]2[/sup] = y(b - x)
a + x[sup]2[/sup] = by - xy
x[sup]2[/sup] + xy - by + a = 0
…so far so good. That gives us fixed values for parameters A through D in (2):
A, B = 1
C, D = 0
E = -b
F = a
Step 2, in which we calculate the hyperbola’s angle of rotation, theta:
The rule I’ve seen (from a reference site ):
if (A == C) theta = pi/4
else theta = 0.5 * atan(B / (A - C));
so:
theta = 0.5 * atan(1) = 0.39269908169872414 radians =~ 22.5 degrees
Given that A, B, and C are constant for all my hyperbolae, the rotation will be constant, too, correct?
Step 3, in which we calculate the coefficients with the rotation removed:
A’ = (A * cos(theta)[sup]2[/sup]) + (B * sin(theta) * cos(theta)) + (C * sin(theta)[sup]2[/sup])
C’ = (A * sin(theta)[sup]2[/sup]) - (B * sin(theta) * cos(theta)) + (C * cos(theta)[sup]2[/sup])
D’ = (D * cos(theta)) + (E * sin(theta))
E’ = (-D * sin(theta)) + (E * cos(theta))
F’ = F
Since we know theta:
sin(theta) = 0.3826834323650898
cos(theta) = 0.9238795325112867
A’ = 1.2071067811865475
C’ = 1.2071067811865475
D’ = -b * sin(theta)
E’ = -b * cos(theta)
F’ = a
At this point, we’ve simply removed the rotation (B) coefficient.
Step 4, in which we calculate the center (x[sub]o[/sub],y[sub]o[/sub]):
x[sub]o[/sub] = -D’/2A’ = -(-b * sin(theta))/(2 * 1.2071067811865475) = (b * sin(theta)) / 0.414213562373095
y[sub]o[/sub] = -E’/2C’ = -(-b * cos(theta))/(2 * 1.2071067811865475) = (b * cos(theta)) / 0.414213562373095
So far, so good? And since we’ve only de-rotated the hyperbola, this is also the center of the original, yes? So, given the center, we should be able to translate the shape such that the new center is (0,0) by simply moving it -x and -y?
Step 5, in which we calculate the co-efficients A" and B" of the hyperbola in standard form:
q = (A’ * x[sub]o[/sub][sup]2[/sup]) + (C’ * y[sub]o[/sub][sup]2[/sup]) - F’
A" = A’/q
C" = C’/q
to reveal the standard conic:
A"x[sup]2[/sup] + C"y[sup]2[/sup] - 1 = 0
**
Step 6**, in which we calculate the standard conic:
So looking to put our equation into the standard form of a hyperbola above (3), we calculate:
a’ [sup]2[/sup] = 1 / min(|A"|,|B"|)
b’ [sup]2[/sup] = 1 / max(|A"|,|B"|)
let c = sqrt(a’ [sup]2[/sup] - b’ [sup]2[/sup])
then if (|A"| < |C"|) foci are at (±c, 0)
else foci are at (0,±c)
and the eccentricity = c / a’
…
At least, I think that’s right. Again, the whole point is to be able to express a’ and b’ from (3) in terms of a and b from (1), externalizing the rotation, translation, and eccentricity into separate variables. I feel confident the rotation will always be the same (~22.5 degrees). I have reason to believe the eccentricity will also be constant for hyperbolae that fit (1), but I’m not able to prove it. I also suspect but am unable to prove that the foci of the standardized hyperbolae will always be on the y-axis. I’m also having trouble visualizing exactly what eccentricity is: the word “skew” comes to mind, but it’s not very precise. My working theory is that, in terms of an affine transform, eccentricity would be analogous to shear in one dimension.