I am interested if there are any "deeper reasons" why the roots of equations of the form
$$ \alpha^{ax^2 + bx + c} = \beta^{dx^2 + fx + g} $$
form hyperbolas on the complex plane. I've managed to prove that it is true using algebra, but want to know if there is any reason aside from a purely algebraic one that this must be true. I am comfortable with the possibility that there's no nice explanation for why and that I'm just searching for meaning where there is none, but as I've yet to study university-level mathematics I was wondering if there is some connection between the topics that I'm yet to learn about.
For some context, I recently had to solve $27^{x^2 - 5} = 9^{2x}$ and came up with the general formula for the roots:
$$ x = \frac{2}{3} \pm \frac{\sqrt{49 + \frac{6\pi i n}{\ln 3}}}{3} $$
for any integer $n$. Out of interest I tried plotting this curve for different values of $n$ and got this shape:
This looks like a hyperbola with the two vertices centered around the two real roots, $3$ and $-5/3$! To prove this, I wrote $x = p + qi$ and then expanded the exponent like so:
$$ \begin{aligned} 27^{x^2 - 5} &= 9^{2x} \\ 27^{(p + qi)^2 - 5} &= 9^{2(p + qi)} \\ 27^{(p^2 - q^2) + 2pqi - 5} &= 9^{2p + 2qi} \\ e^{\ln(27) ((p^2 - q^2) + 2pqi - 5)} &= e^{\ln(9) (2p + 2qi)} \\ e^{\ln(27) ((p^2 - q^2) + 2pqi - 5) - \ln(9) (2p + 2qi)} &= 1 \\ \ln(27)((p^2 - q^2) + 2pqi - 5) - \ln(9) (2p + 2qi) &= 2\pi i n \end{aligned} $$
Equating real and imaginary parts gives:
$$ \ln(27)(p^2 - q^2 - 5) - 2\ln(9)p = 0 \\ 2\ln(27)pqi - 2\ln(9)qi = 2\pi i n $$
This can be further rearranged to show that all the roots lie at the intersection of one hyperbola and an infinite family of additional hyperbolas at right angles to the initial one:
$$ \left(p-\frac{2}{3}\right)^2 - q^2 = \frac{49}{9} \\ \left(p-\frac{2}{3}\right)q = \frac{\pi n}{\ln(27)} $$
Here is a plot of the above two curves. The red hyperbola represents the first equation and the light green hyperbolas represent the infinite family specified by the second equation:
A similar argument can be made for any equation $\alpha^{ax^2 + bx + c} = \beta^{dx^2 + fx + g}$, and for the case $e^{x^2 - 1} = 1$ then the roots all lie on the unit hyperbola, $x^2 - y^2 = 1$.
Another method of proof is to show that separating the function for the roots in terms of the parameter $n$ into a function for the real and imaginary part of the $x$ satisfies the equation $\frac{(\Re(x) - j)^2}{k} - \frac{\Im(x)^2}{l} = 1$ for some arbitrary constants $j$, $k$ and $l$.
Again, my question is: are there any "deeper reasons" that this is true? Is there a connection between roots to exponential equations like this and conic sections? Or even, is there a geometrical argument that can be made here? Or is it that this algebraic relationship is the connection between the two, and that I'm looking for meaning when there actually is none.
Thanks in advance for any replies.