I have a rational equation derived from 2 points, $(2, 2)$ and $(10, 10)$. Solving for the rational equation gives the equation $$y = \frac{20}{12-x}.$$

What I want to happen right now is that given a new point $(x, y)$, and assuming that the new points $x$ and $y$ won't make the graph linear or look like a logarithmic graph, I'd like to know the new equation.

So far, sticking with the formula for a rational function $$y = \frac{m}{c-x}$$ doesn't help as much since I have $3$ points but only $2$ unknowns to work with. My idea right now is to solve look for an ellipse whose axes are fixed at $(2, 2)$ and $(10, 10)$. From there, I can start to figure out if I can plug in new $x$ and $y$ values to adjust the curvature of the ellipse while still holding $(2, 2)$ and $(10, 10)$ at position. Obviously the center won't be at the origin any longer.

Can anyone help me with my issue? I'm really rusty with my conic section maths.

Update Edit:

As I'm looking more and more onto this problem, the more I realise that I can't tell for sure if I need a rational equation or an equation for a conic section.

A rational equation wouldn't do since it only takes two points to determine a graph, adding a third point complicate things as I now have an overdetermined system of equations.

Using a conic section (ellipse) might be and issue as well since the center may move and other constants such as foci length will adjust as well when another point needs to be taken into account and the whole ellipse adjusts.

I also discussed this with a friend of mine, and she recommended that I use a third degree polynomial or a monotonic curve. I will look into that.

2nd Update:

A Polynomial of the n-1th degree given n points does not work. Given our example before where the boundaries $(2, 2)$ and $(10, 10)$, if the point is set nicely between the two boundaries, then I can come up with a system of 3 equations and Gauss-Jordan Elimination solves that nicely.

However, in the case that the third point is $(9, 4)$, which is a bit too low, what happens is that the parabola is shown where the $(2, 2)$ point is at the left part of the parabola and the two other points are in the right part. Adding a fourth point would make the graph curve out more than needed.

The reason why I wanted to find out if there is a way to find an upward-facing curve given two boundary points and a third setting point is because we wanted to make a flexible graph wherein anyone can set a third point and the graph would adjust accordingly.

However, it seems that there is no clean mathematical way to go about it. I found a solution Bezier Curve that fits well with what I wanted to achieve.

However, I am still interested if anyone manages to find out a way to come up with an upward facing curve given a certain number of points.

  • $\begingroup$ Well, as you said, you have two unknowns and three constraints. So odds are that you can't find such a solution. For a general conic (which includes an ellipse), it takes five points to uniquely determine a conic. It takes three to determine a circle, however. I guess the real question is, why are you trying to do this? What particular shape do you want the graph to have, and why? $\endgroup$ – Simon Rose Jan 5 '15 at 9:38
  • $\begingroup$ Well, we wanted to determine a graph that would show what discount to give given a certain quantity, with the initial rational function as a "baseline". Now, we wanted to be able to "adjust" the graphs curve given a third point (x, y) that's between the initial bounds, while holding the initial bounds (2,2) and (10, 10) in place. $\endgroup$ – Razgriz Jan 5 '15 at 9:43
  • $\begingroup$ I see. The issue with this in some sense is that there are infinitely many functions which pass through (2,2) and (10,10) and any other point you care about. Without having some a priori reason to pick one form or another, you have no real way to uniquely determine an answer. $\endgroup$ – Simon Rose Jan 5 '15 at 10:09
  • $\begingroup$ I understand. We tried doing it with polynomials but to no avail. Actually we found a more programming-aligned solution recently. $\endgroup$ – Razgriz Jan 5 '15 at 11:33

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