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.

Solution 1:

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 $$ y={\pm\sqrt{1+a^2}-1\over a}z. $$