J. Mike Rollins (Sparky) [rollins@wfu.edu]
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Given 3 points, draw a hyperbola.

Yesterday, I needed to generate the equation for a hyperbola given three points. I still had a spread sheet from a few years ago when I had to do this, but I did not have any of the derivation. So, I figured I would write this up nicely and put this on the web.

For example, what is the equation of a hyperbola that passes through the points (25,1), (55,0.5), and (165,0) where the equation is of the form Y=A/(X-B)+C?

The equation Y=A/(X-B)+C can be worked into the following form.

This form of the equation is a little easier to work with when solving three equations with three unknown. We will use Alpha for our X values and Beta for your Y values. Here are the three equations we will use.

I observe that it is easy to eliminate the variable A. So, I take the first two equations and subtract. I do the same with the first and third equation.

Each equation can be factored to make it a little more manageable.

The two previous equations are in terms of B and C. We use those two equations and eliminate C. (I originally meant to eliminate B so it would look a little nicer...O well!)

I used division to create the common coefficient for the term C.

Now we have a single equation with a single unknown, B.

We will now solve for B.

We can now solve for C.

Finally, we solve for A.

In summary: The hyperbola of the form Y=A/(X-B)+C that passes through the points (a1,b1), (a2,b2) and (a3,b3), has values for A, B and C as follows: