# Locating the vertex of a hyperbola from a particular function

I've made a function for my physics research. My advisor wants all of the parameters to have a clear physical meaning. I've been calling one of the parameters "transition interval" but a hyperbola doesn't have a parameter or characteristic by such a name. The latus rectum seems to be a characteristic that is related to that idea. I'm pretty sure the function is a hyperbola, but it doesn't have the characteristics I'm expecting. $$f(x)$$ is unitless, $$a$$ is inverse energy, and $$x_0$$, $$x$$, and $$b$$ are energy. $$f(x):=\frac{a}{2}(x-x_0-\sqrt{b^2+(x-x_0)^2})$$ Due to units getting in the way of analytic geometry, I've modified the function so that f(z) and z are unitless. $$f(z):=\frac{1}{2}(z-a x_0-\sqrt{a^2 b^2+(z-a x_0)^2})$$ I'm pretty sure both of these are hyperbolas. Mathematica says it can get both into a form of $$A_{xx}x^2+2A_{xy}xy+A_{yy}y^2+2B_xx+2B_yy+C=0$$. The left side goes off to a linear asymptote of $$z-a x_0$$ and the right side goes off to a linear asymptote of $$0$$. The asymptotes intersect at the center of $$(a x_0,0)$$. The primary axes are $$1/2(z-a x_0)$$ and $$-2(z-a x_0)$$.

This is where things go funny. The first axis should be parallel to a tangent line that passes through the vertex. The second axis should intersect the hyperbola at the vertex. Under the first characteristic of the vertex I find the vertex is at $$(a x_0,-\frac{1}{2}a b)$$. Under the second characteristic of the vertex I find the vertex is at $$(\frac{1}{12}(\sqrt{6}a b+12 a x_0),-\frac{a b}{\sqrt{6}})$$. There can only be one vertex. Once I have a vertex, and by extension the semi-major axis, finding the semi-minor axis, the focus, and the latus rectum is straight forward for me.

From assuming the second axis intersecting the hyperbola defines the vertex, I've found the semi-major axis is $$\frac{1}{2}\sqrt{\frac{5}{6}}a b$$, the semi-minor axis is $$\sqrt{\frac{5}{6}}a b$$, the eccentricity is $$\sqrt{5}$$, the focus is at $$(\frac{1}{2}\sqrt{\frac{5}{6}}a b +a x_0,-\sqrt{\frac{5}{6}} a b)$$, and the latus rectum intersects the hyperbola at $$(\frac{1}{12}(-\sqrt{606} a b + 12 a x_0), \frac{1}{24}(-5 \sqrt{30} a b -\sqrt{606}a b))$$ and $$({\frac{1}{12}(\sqrt{606}a b + 12 a x_0), \frac{1}{24}(-5\sqrt{30} a b +\sqrt{606}a b)})$$. It was at this point that I noticed that one point is $$\frac{1}{4}\sqrt{\frac{5}{3}(53+\sqrt{505})} a b$$ from the focus and the other is $$\frac{1}{4}\sqrt{\frac{5}{3}(53-\sqrt{505})} a b$$ from the focus and those should be the same and I came here. So either 1) this hyperbola has axes which aren't perpendicular to each other, 2) it isn't symmetric across the major axis, or 3) it isn't a hyperbola and when Mathematica says a function fits the form $$A_{xx}x^2+2A_{xy}xy+A_{yy}y^2+2B_xx+2B_yy+C=0$$ it may be lying to me. Any of these would explain why Wikipedia's article on hyperbolas didn't apply this function; the semi-minor axis provided in "2.11 Quadratic Equation" was imaginary.

• I used the equation y^2+a*h*y-a*x*y-(a*b/2)^2=0 from the first y=f(x) and I get the foci to be {(h-(a^2+1)^(1/4)*cos(atan2(a,-1)/2)*b, (a^2+1)^(1/4)*sin(atan2(a,-1)/2)*b), (h+(a^2+1)^(1/4)*cos(atan2(a,-1)/2)*b,-(a^2+1)^(1/4)*sin(atan2(a,-1)/2)*b)} The asymptotes to be y*(y-a*x+a*h)=0 The axes to be a*y^2+2*x*y-2*h*y-a*x^2+2*a*h*x-a*h^2=0 Commented Jan 1, 2022 at 11:47
• I got the solution to my immediate problem. Its going to take a bit more time to work out the result. I'll give my answer when I've finished with it. But it is too late right now to finish it. @Jan-MagnusØkland, I don't know where that 'h' came from. Commented Jan 2, 2022 at 8:05
• h is my name for $x_0.$ Commented Jan 2, 2022 at 8:14

Relabelling $$z=x-x_0$$ and $$y=f(z)$$ your first equation reads $$y={a\over2}\big(z-\sqrt{z^2+b^2}\big).$$ Rearranging and squaring yields $$y^2-ayz={a^2b^2\over4},$$ that is: $$y(y-az)={a^2b^2\over4}.$$ This is the equation of a hyperbola with asymptotes $$y=0$$ and $$y=az$$, confirming what you found. But the axes are the angle bisectors of the asymptotes, hence their equations you cited in the question cannot be correct. The right result for the equation of the axes is $$y={\pm\sqrt{1+a^2}-1\over a}z.$$

• Careful. "a" has units and 1 does not. You should relabel z=a(x-x0). That would make z unitless as well instead of inverse units of "a." For analytic geometry to work well all dimensions must have the same units. Commented Jan 1, 2022 at 23:58
• An angle bisector is not a perpendicular bisector. That is, a slope of 1/2 does not bisect a slope of 1, but an angle of pi/8 radians does bisect an angle of pi/4 and a slope of 1. Commented Jan 2, 2022 at 7:16
• @user2603428 You are right: there was an error in the equations of the axes, which I have corrected. You can revert to your notations plugging back $z=(x-x_0)$. Units are not relevant here: I can regard $y$, $z$ and $b$ as lengths, and $a$ as an dimensionless parameter, as long as I'm interested in their geometrical meaning. Of course they become relevant when you apply the equations to a physical problem. Commented Jan 2, 2022 at 8:24
• The second comment was about what I had wrong in getting to my bad answer. I just really have a hard time with (sqrt(1+(5GeV^-1)^2)-1)/(5GeV^-1) and would rather not have to figure out what it is supposed to mean when I convert back to physics. Commented Jan 2, 2022 at 22:54

Thanks to @Intelligenti pauca for pointing to my error and how to get to the answer that I find satisfactory.

First I redefine f(x) such that it the independent and dependent variables are unitless. Analytic geometry requires that all dimensions have the same unit for anything to be sensible. What is the length of the major and minor axes in a rotated frame supposed to be? What does $$\sqrt{(5\text{GeV})^2+3^2}$$ mean? It doesn't mean anything. So best to keep things consistent. $$z:=a(x-x_0)$$ $$f(z):=\frac{1}{2}(z-\sqrt{a^2b^2+z^2})$$ This gives me an asymptote of $$y=z$$ and $$y=0$$. The major and minor axes bisect the angles, not what would be the perpendicular bisector of a triangle. That gives axes of $$y=z\tan(\frac{\pi}{8})$$ and $$y=-z\cot(\frac{\pi}{8})$$ or $$y=a(x-x_0)\tan(\frac{\pi}{8})$$ and $$y=-a(x-x_0)\cot(\frac{\pi}{8})$$ in the physical space, which don't appear to be perpendicular. But what is perpendicular in the physical space? Two lines that are $$\pi/4$$ radians apart. What is a radian? The angle where the radius is equal to the arclength. What is the radius of a circle where the length is 1 GeV in one direction and 1 in the other direction? That clearly isn't a sensible question. This will lead to funny results in the physical space. The function in the physical space is also clearly a hyperbola with it's own perpendicular axes. Using those axes will be consistent with the answer given by @Intelligenti pauca, but the answer will have unit conflicts.

Now that I have axes, I can run the major axis to the intersection with the hyperbola. That gives me a vertex at $$\left(-\frac{2 \sqrt[4]{2} a b \sin \left(\frac{\pi }{8}\right)}{\sqrt{2-\sqrt{2}}-\sqrt{1+\cot \left(\frac{\pi }{8}\right)}},-\frac{2 \sqrt[4]{2} a b \sin \left(\frac{\pi }{8}\right)}{\sqrt{2-\sqrt{2}}-\sqrt{1+\cot \left(\frac{\pi }{8}\right)}}\right)$$. I would have been happy with this answer, as I could take it and run with it to the actual answer that I needed. Having found the answer final answer that I need, I can't keep it to myself, so here's the plot of everything in the unitless space.

The z coordinates of the primary latus rectum are $$z_1=-\frac{a b \csc \left(\frac{\pi }{8}\right) \left(8-4 \cot \left(\frac{\pi }{8}\right)+\sqrt[4]{2} \sqrt{3 \sqrt{2}+4} \csc \left(\frac{\pi }{8}\right)\right)}{4\ 2^{3/4} \left(\cot \left(\frac{\pi }{8}\right)-1\right)}\approx-1.99546ab$$ $$z_2=\frac{2 \sqrt[4]{2} ab \sin \left(\frac{\pi }{8}\right)}{\sqrt{2-\sqrt{2}}-\sqrt{1+\cot \left(\frac{\pi }{8}\right)}}\approx2.90564ab$$

Now back in the physical space, the vertex is located at $$\left(\frac{1}{2} \sqrt{\frac{3}{\sqrt{2}}-2} b+x_0,-\frac{1}{8} a \left(\sqrt{6 \sqrt{2}+8} b-\sqrt{6 \sqrt{2}-8} b\right)\right)$$. For $$a=5\text{GeV}^{-1}$$, $$b=.75\text{GeV}$$, and $$x_0=2.5\text{GeV}$$, that works out to $$\left(\left(\frac{3}{8} \sqrt{\frac{3}{\sqrt{2}}-2}+\frac{5}{2}\right)\text{GeV},-\frac{15}{8 \sqrt[4]{2}}\right)\approx(2.63062\text{GeV},-1.57668)$$. The end points for that latus rectum are approximately $$(1.0034\text{GeV},-7.92651)$$ and $$(4.67923\text{GeV},-.313621)$$. How long is that latus rectum? No idea, that isn't a sensible question. But how long is the x interval is a sensible question, and the complete answer I was actually after. I thought it would be discourteous to ask this forum for the complete answer. That interval is$$\frac{\sqrt{17 \sqrt{2}+24}}{\cot \left(\frac{\pi }{8}\right)-1}b\approx4.9011b$$.