# Tag Archives: conics

## CAS Presentations at USACAS-9

I had two presentations at last Saturday’s USACAS-9 conference at Hawken School in Cleveland, OH.  Following are outline descriptions of the two sessions with links to the PowerPoint, pdf, and .tns files I used.  I’m also adding all of this information to the Conference Presentations tab of this ‘blog.

Powerful Student Proofs

This session started with a brief introduction to a lab that first caught my eye at the first USACAS conference years ago.

You know how the graph of $y=ax^2+bx+c$ behaves when you vary a and c, but what happens when you change b?

I ‘blogged on this problem here and here.  In the session, we used TI-Nspire file QuadExplore.

Next, we explored briefly the same review of trigonometric and polar graphs not as static parent functions under static transformations, but as dynamic curves oscillating between their ceilings and floors.  In the session, we used TI-Nspire file Intro Polar.

Having a complete grasp of polar graphs of limacons, cardioids, rose curves, and hybrids of these, I investigated what would happen for curves of the family $r=cos \left( \frac{\theta}{k} \right)$.  Curiously, for $k=3$, I encountered a curve that looked like a horizontal translation of limacons–something that just shouldn’t happen within polar coordinates.

One of my former students, Sara, used a CAS to convert a polar curve to Cartesian, translate the curve, and convert back to polar.  She then identified and solved a trig identity to confirm what the graph suggested.  A complete description of Sara’s proof is below.  I originally ‘blogged on Sara’s work here which was a much more elegant solution to the problem than my initial attempt.  It’s always cool when a student’s work is better than her teacher’s!  I used TI-Nspire file Polar Fractions in Saturday’s session.

The last example presented itself when I created a document to model the family of conic curves resulting from manipulating the coefficients of $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$.  After I created  dynamic points for the foci of the conics, something unusual happened when the E parameter for horizontal ellipses and hyperbolas varied.

The foci for hyperbolas followed an ellipse, and the locus of elliptical foci appeared to be a hyperbola.  Another former student, Lilly, proved this property to be true.  A detailed explanation of Lilly’s proof is below.  We were fortunate to have Lilly’s work published in the Mathematics Teacher in May, 2014.

To demonstrate this final part of the session, I used TI-Nspire file Hidden Conic Behavior.

Here is my PowerPoint file for Powerful Student Proofs.  A more detailed sketch of the session and the student proofs is below.

Bending Asymptotes, Bouncing Off Infinity, and Going Beyond

The basic proposal was that adding the Reciprocal transformation to the palette of constant dilations and translations dramatically simplified understanding of the behavior of rational functions around even and odd vertical asymptotes (bouncing off and passing through infinity).  Just like lead coefficients of polynomials determine their end behavior, so, too, do the lead coefficients of proper rational expressions define the end behavior of rational functions.

Extending the idea of reciprocating and transforming functions, you can quickly explain exponential decay from exponential growth, derive the graphs of $y=\frac{1}{x}$ and $y=\frac{1}{x^2}$, and completely explain why logistic functions behave the way they do.

We finished with a quick exploration of trigonometric and polar graphs not as static parent functions under static transformations, but as dynamic curves oscillating between their ceilings and floors.

I used TI-Nspire Bending and Intro Polar files in the demonstration.  Here is my outline PowerPoint file for Bending Asymptotes.

## Cover Article

I was pretty excited yesterday when the latest issue of NCTM’s Mathematics Teacher arrived in the mail and the cover story was an article I co-wrote with a former student who’s now at MIT.

The topic was the finding and proof of a cool interconnected property of the foci of hyperbolas and ellipses that I made years ago when setting up my TI-Nspire CAS to model conic sections via the polynomial definition.

After pitching the idea to teachers at professional conferences for a couple years with no response, I asked one of my 9th grade students if she’d be interested in a challenge.  Her eventual proof paralleled mine, and our work together enhanced and polished each other’s understanding and proofs.

While all of the initial work was done with the TI-Nspire CAS, we wrote the article using GeoGebra so that readers could freely access Web-based documents to explore the mathematics for themselves.

You can access the article on the NCTM site here.

While a few minor changes happened after it was created, here is a pre-publication proof of the article.

## Traveling Dots, Parabolas, and Elegant Math

Toward the end of last week, I read a description a variation on a paper-folding strategy to create parabolas.  Paraphrased, it said:

1. On a piece of wax paper, use a pen to draw a line near one edge.  (I used a Sharpie on regular copy paper and got enough ink bleed that I’m convinced any standard copy or notebook paper will do.  I don’t think the expense of wax paper is required!)
2. All along the line, place additional dots 0.5 to 1 inch apart.
3. Finally, draw a point F between 0.5 and 2 inches from the line roughly along the midline of the paper toward the center of the paper.
4. Fold the paper over so one of the dots on line is on tope of point F.  Crease the paper along the fold and open the paper back up.
5. Repeat step 4 for every dot you drew in step 2.
6. All of the creases from steps 4 & 5 outline a curve.  Trace that curve to see a parabola.

I’d seen and done this before, I had too passively trusted that the procedure must have been true just because the resulting curve “looked like a parabola.”  I read the proof some time ago, but I consumed it too quickly and didn’t remember it when I was read the above procedure.  I shamefully admitted to myself that I was doing exactly what we insist our students NEVER do–blindly accepting a “truth” based on its appearance.  So I spent part of that afternoon thinking about how to understand completely what was going on here.

What follows is the chronological redevelopment of my chain of reasoning for this activity, hopefully showing others that the prettiest explanations rarely occur without effort, time, and refinement.  At the end of this post, I offer what I think is an even smoother version of the activity, freed from some of what I consider overly structured instructions above.

CONIC DEFINITION AND WHAT WASN’T OBVIOUS TO ME

A parabola is the locus of points equidistant from a given  point (focus) and line (directrix).

What makes the parabola interesting, in my opinion, is the interplay between the distance from a line (always perpendicular to some point C on the directrix) and the focus point (theoretically could point in any direction like a radius from a circle center).

What initially bothered me about the paper folding approach last week was that it focused entirely on perpendicular bisectors of the Focus-to-C segment (using the image above).  It was not immediately obvious to me at all that perpendicular bisectors of the Focus-to-C segment were 100% logically equivalent to the parabola’s definition.

1. I knew without a doubt that all parabolas are similar (there is a one-to-one mapping between every single point on any parabola and every single point on any other parabola), so I didn’t need to prove lots of cases.  Instead, I focused on the simplest version of a parabola (from my perspective), knowing that whatever I proved from that example was true for all parabolas.
2. I am quite comfortable with my algebra, geometry, and technology skills.  Being able to wield a wide range of powerful exploration tools means I’m rarely intimidated by problems–even those I don’t initially understand.  I have the patience to persevere through lots of data and explorations until I find patterns and eventually solutions.

I love to understand ideas from multiple perspectives, so I rarely quit with my initial solution.  Perseverance helps me re-phrase ideas and exploring them from alternative perspectives until I find prettier ways of understanding.

In my opinion, it is precisely this willingness to play, persevere, and explore that formalized education is broadly failing to instill in students and teachers.  “What if?” is the most brilliant question, and the one we sadly forget to ask often enough.

ALGEBRAIC PROOF

While I’m comfortable handling math in almost any representation, my mind most often jumps to algebraic perspectives first.  My first inclination was a coordinate proof.

PROOF 1:  As all parabolas are similar, it was enough to use a single, upward facing parabola with its vertex at the origin.  I placed the focus at $(0,f)$, making the directrix the line $y=-f$.  If any point on the parabola was $(x_0,y_0)$, then a point C on the directrix was at $(x_0,-f)$.

From the parabola’s definition, the distance from the focus to P was identical to the length of CP:

$\sqrt{(x_0-0)^2-(y_0-f)^2}=y_0+f$

Squaring and combining common terms gives

$x_0 ^2+y_0 ^2-2y_0f+f^2=y_0 ^2+2y_0f+f^2$
$x_0 ^2=4fy$

But the construction above made lines (creases) on the perpendicular bisector of the focus-to-C segment.  This segment has midpoint $\displaystyle \left( \frac{x_0}{2},0 \right)$ and slope $\displaystyle -\frac{2f}{x_0}$, so an equation for its perpendicular bisector is $\displaystyle y=\frac{x_0}{2f} \left( x-\frac{x_0}{2} \right)$.

Finding the point of intersection of the perpendicular bisector with the parabola involves solving a system of equations.

$\displaystyle y=\frac{x_0}{2f} \left( x-\frac{x_0}{2} \right)=\frac{x^2}{4f}$
$\displaystyle \frac{1}{4f} \left( x^2-2x_0x+x_0 ^2 \right) =0$
$\displaystyle \frac{1}{4f} \left( x-x_0 \right) ^2 =0$

So the only point where the line and parabola meet is at $\displaystyle x=x_0$–the very same point named by the parabola’s definition.  QED

Proof 2:  All of this could have been brilliantly handled on a CAS to save time and avoid the manipulations.

Notice that the y-coordinate of the final solution line is the same $y_0$ from above.

MORE ELEGANT GEOMETRIC PROOFS

I had a proof, but the algebra seemed more than necessary.  Surely there was a cleaner approach.

In the image above, F is the focus, and I is a point on the parabola.  If D is the midpoint of $\overline{FC}$, can I conclude $\overline{ID} \perp \overline{FC}$, proving that the perpendicular bisector of $\overline{FC}$ always intersects the parabola?

PROOF 3:  The definition of the parabola gives $\overline{FI} \cong \overline{IC}$, and the midpoint gives $\overline{FD} \cong \overline{DC}$.  Because $\overline{ID}$ is self-congruent, $\Delta IDF \cong \Delta IDC$ by SSS, and corresponding parts make the supplementary $\angle IDF \cong \angle IDC$, so both must be right angles.  QED

PROOF 4:  Nice enough, but it still felt a little complicated.  I put the problem away to have dinner with my daughters and when I came back, I was able to see the construction not as two congruent triangles, but as the single isosceles $\Delta FIC$ with base $\overline{FC}$.  In isosceles triangles, altitudes and medians coincide, automatically making $\overline{ID}$ the perpendicular bisector of $\overline{FC}$.  QED

Admittedly, Proof 4 ultimately relies on the results of Proof 3, but the higher-level isosceles connection felt much more elegant.  I was satisfied.

TWO DYNAMIC GEOMETRY SOFTWARE VARIATIONS

Thinking how I could prompt students along this path, I first considered a trace on the perpendicular lines from the initial procedure above (actually tangent lines to the parabola) using to trace the parabolas.  A video is below, and the Geogebra file is here.

http://vimeo.com/89759785

It is a lovely approach, and I particularly love the way the parabola appears as a digital form of “string art.”  Still, I think it requires some additional thinking for users to believe the approach really does adhere to the parabola’s definition.

I created a second version allowing users to set the location of the focus on the positive y-axis and using  a slider to determine the distances and constructs the parabola through the definition of the parabola.  [In the GeoGebra worksheet (here), you can turn on the hidden circle and lines to see how I constructed it.]  A video shows the symmetric points traced out as you drag the distance slider.

A SIMPLIFIED PAPER PROCEDURE

Throughout this process, I realized that the location and spacing of the initial points on the directrix was irrelevant.  Creating the software versions of the problem helped me realize that if I could fold a point on the directrix to the focus, why not reverse the process and fold F to the directrix?  In fact, I could fold the paper so that F touched anywhere on the directrix and it would work.  So, here is the simplest version I could develop for the paper version.

1. Use a straightedge and a Sharpie or thin marker to draw a line near the edge of a piece of paper.
2. Place a point F roughly above the middle of the line toward the center of the paper.
3. Fold the paper over so point F is on the line from step 1 and crease the paper along the fold.
4. Open the paper back up and repeat step 3 several more times with F touching other parts of the step 1 line.
5. All of the creases from steps 3 & 4 outline a curve.  Trace that curve to see a parabola.

This procedure works because you can fold the focus onto the directrix anywhere you like and the resulting crease will be tangent to the parabola defined by the directrix and focus.  By allowing the focus to “Travel along the Directrix”, you create the parabola’s locus.  Quite elegant, I thought.

As I was playing with the different ways to create the parabola and thinking about the interplay between the two distances in the parabola’s definition, I wondered about the potential positions of the distance segments.

1. What is the shortest length of segment CP and where could it be located at that length?  What is the longest length of segment CP and where could it be located at that length?
2. Obviously, point C can be anywhere along the directrix.  While the focus-to-P segment is theoretically free to rotate in any direction, the parabola definition makes that seem not practically possible.  So, through what size angle is the focus-to-P segment practically able to rotate?
3. Assuming a horizontal directrix, what is the maximum slope the focus-to-P segment can achieve?
4. Can you develop a single solution to questions 2 and 3 that doesn’t require any computations or constructions?

CONCLUSIONS

I fully realize that none of this is new mathematics, but I enjoyed the walk through pure mathematics and the enjoyment of developing ever simpler and more elegant solutions to the problem.  In the end, I now have a deeper and richer understanding of parabolas, and that was certainly worth the journey.

## Rotating Parabolas

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 $y=a\cdot x^2+b\cdot x+c$, 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 $\theta$ units contained those points, then if I could rotate the points back $\theta$ 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, $a\cdot x^2+b\cdot x+c$ as shown below.

Those bottom three lines definitely look intimidating, but for any constant value of $\theta$, 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 $\theta$ for a, b, and c back into the standard quadratic, $y=a\cdot x^2+b\cdot x+c$ 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 $\theta$ 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 $0\leq\theta\leq 2\pi$.

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.

## March 2011 Conference Presentations

T3 Regional Conference – Suwanee, GA – Saturday, March 19, 2011.

PreCalculus:  Transformed & Nspired

This workshop offers an innovative understanding of pre-calculus concepts through nonstandard transformations, allowing functions and concepts to be unified by a handful of underlying mathematical structures. It provides approaches that dramatically simplify many initially complicated-looking problems. CAS-enhanced ideas are presented.  (Co-presented with Nurfatimah Merchant)

Conics within Conics

This session presents the family of conic sections by connecting their algebraic and graphical representations, showing how each section can evolve from the others. The conclusion is a surprisingly elegant conic property and a 9th grader’s proof submitted for publication.