When one graphs rational functions in Pre-Calculus type course, one usually graphs functions that are reciprocals of linear functions and reciprocals of quadratics. One of the other properties that maybe asked is to find the invariant points. Sometimes the invariant points correspond to the center of curvature of the arms of the hyperbolas that are created from these functions.
Is there a way to determine the center of curvature for the hyperbola arms of these rational functions without using Calculus based methods.
When i say center of curvature that means i am looking for the coordinates(x,y) of the center.
To make this more clear I have added an image i generated:
As you can see, I put the quadratic as well to get intersection with the reciprocal which gives the invariant points. I thought I would try to add a line that connect the peak of the bottom quadratic like parabola with vertical asymptote, but it went lower than the center of curvature of the hyperbolic arm. (The line is in green color)
I just looked at the graph and decided to try to put a line thru the asymptote at x =2 to with a slope of 1, and it looks like its going thru the center of curvature of the hyperbolic arm.
See attached image, the new line is in orange color:
Also wish to add further analogy to the actual conic section Hyperbola, in this the center is called the Vertex of the Hyperbola. See attached image:
I also want to note something I just noticed, that the line that goes thru the hyperbolic arms of this reciprocal quadratic, cuts the arms in a very symmetrical way such that the line acts like a mirror, cutting it symmetrically in half.