Here’s a fun problem that continues to grow that Nurfatimah Merchant and I included in our textbook .

**How many uniquely defined curves can you find whose graphs contain the points (1,1), (6,-3), and (7,3)?**

NOTE: Some of the algebra below is very intimidating to those who aren’t pretty good friends with mathematical symbol-pushing. If you like, you can skip to the brief video clip at the end of this post showing all solutions.

**SOLUTION ALERT — Don’t read any further if you want to play with this problem yourself.**

Many students instantly think of vertical parabolas of the form , but when I presented this at the MMC meeting in Chicago tonight, two stellar Geometry teachers on the front row suggested circles. Others correctly noted that there are infinitely many curves defined solely by those three points, but I’ll talk about that in a second. Most students first think about vertical parabolas and circles.

From the generic equation for a vertical parabola, one could plug in the three given ordered pairs and create a 3×3 system of linear equations. I argue that a CAS is a good call here, especially when it can give the quadratic equation in multiple forms with equal ease.

That’s lots of algebra, but it was all completed very quickly using my CAS, proving another technology advantage is that all forms of an equation are equally easy to compute on a CAS. When people try to solve this problem by hand, they invariably use the standard form only because the algebraic manipulations in the other cases are unwieldy, at best. Here’s an image of the curve using the factored form of the equation.

The circle was also easy to get using a CAS to drive the algebra.

This graph shows the circle and the vertical parabola together.

A horizontal parabola can be obtained just as easily, but I’ll leave that one for you to discover.

This past summer, I was prodded by a teacher at the institute Nurfatimah and I ran for Westminster’s Center for Teaching to find all rotated parabolas which contained those points. To do so, I thought that if a parabola rotated some units contained those points, then if I could rotate the points back units, the corresponding parabola would be vertical, and I could solve it using my CAS as above.

Using a transformation matrix on my original 3 points, I found their rotation images and substituted these into the standard form of a vertical parabola, as shown below.

Those bottom three lines definitely look intimidating, but for any constant value of , each is just a linear equation in a, b, and c. While my Nspire CAS couldn’t solve that system directly and WolframAlpha timed out, I could accomplish the algebra step-by step on the Nspire. (Click here if you have TI-Nspire CAS software and want to see my algebra.)

Geogebra 4.2 Beta did accomplish the solution in a single line. Its solution to each coefficient is shown here:

Substituting the expressions in terms of for a, b, and c back into the standard quadratic, and rotating the equation back into place created a generic equation for a rotatable parabola through the given three points. I used this equation to define my rotatable parabola through the 3 points with a slider for here on GeoGebraTube (here is the original GeoGebra document). You don’t get to manipulate it, but the following vimeo clip shows all of the rotated parabolas for .

As a final bang for the presentation, one teacher in the audience wondered what it would look like if a trace was placed on the rotating parabolas. Easily done. Whether doing this in an Nspire CAS or on GeoGebra, right click the curve and select trace on. Dragging the slider for the angle through all of its possible values creates the following graph.

Now that’s just pretty! The enveloped triangle in the center is precisely the triangle circumscribed by the circle found above. It’s straight edges are defined by the degenerate cases of the parabolas at the instances when the vertices were stretched to infinity.

As some in tonight’s audience pointed out, there are also infinitely many rotated circles and ellipses defined by these points. Another wondered what would happen if a trace was placed on the rotating vertex instead of the entire parabola. On that last question, we’re pretty certain that the curve is some sort of “tri-perbola”–a ‘hyperbola’ with three branches–whose vertices are the three given points. We don’t know an equation for it (yet), so that and other great extensions of this are now problems for another day. I’d love to hear what others think or find in this problem.