This past spring (of 2020), when my university in Istanbul was closed (like all others in Turkey) against the spread of the novel coronavirus, I created for my students an exercise, to serve at least as a distraction for those who could find distraction in learning.
Note added, April 17, 2023: An account of the mathematics involved in the exercise would ultimately be published as: Pierce, D. (2021). “Conics in Place.” Annales Universitatis Paedagogicae Cracoviensis | Studia Ad Didacticam Mathematicae Pertinentia, 13, 127–150.
The exercise uses no more mathematical tools than may be found in an algebra course in high school; yet it serves the purposes of university mathematics, as I understand them.
My course was in analytic geometry, and our first-year students take this course in their second semester, after a first-semester course of reading and presenting the propositions of Book I of Euclid’s Elements. (The students also have courses in calculus, number theory, and linear algebra in their first year.)
The writings and instructions that I gave to my students are gathered on a page that I called “Analitik Geometri Özeti” (Summary of Analytic Geometry). That is what at first I expected the page (and linked pages) to be: a précis, both of what I had already lectured on, the few times I had met the students in person, and of what I would have gone on to lecture on.
Then I conceived the exercise that shall describe here. I am also posting, as a pdf file, a worked example, in English. (As of August 25, 2020, the linked document has 31 pages of size A5, including 14 figures. This is a shortening of the document that I originally posted, which has 35 pages and 15 figures. The old document is not in error, but I have seen a way to simplify some computations.)
The exercise illustrates the correspondence between conic sections and second-degree equations in two variables.
This correspondence is worked out in some books, such as M. Vygodsky, Mathematical Handbook: Higher Mathematics (Moscow: Mir, Fifth printing 1987), where there is a thorough treatment. However, here and in other sources, the correspondence is worked out by means of changes of coordinates. I use other means, namely the tools of high-school algebra, as I said.
Moreover, I do not contrive examples that will work out neatly. I ask students to supply their own examples.
Briefly, the exercise is to
- choose three points in the coordinate plane,
- use these to define a central conic section, namely an ellipse or hyperbola,
- find the axes of that conic.
Strictly, the first two points are not that, but vectors, determining conjugate diameters of a conic section whose center is the third point.
One can plot one’s results on graph paper, by way of confirming that one’s computations are correct.
There could be a similar exercise involving the parabola.
Perhaps every course of analytic geometry will teach the ellipse as being given by an equation of the form
(x/a)2 + (y/b)2 = 1;
the hyperbola,
(x/a)2 − (y/b)2 = 1.
These are central conics, their centers being the origin. The conics have axes, one with endpoints (±a, 0); the other, (0, ±b). The latter segment, since it does not actually meet the hyperbola, may not normally be considered an axis of the hyperbola. We shall nonetheless call it that, following Apollonius, and noting that the two axes are shared with the conjugate hyperbola, given by
(y/b)2 − (x/a)2 = 1.
Apollonius proves that every chord of a central conic that passes through the center is a diameter and has a conjugate diameter. Vygodsky states the result without proof; it means that chords of the conic that are parallel to one of the diameters are bisected by the other.
In another post, “Elliptical Affinity,” I tried to make the Apollonian proof visual. The exercise that I am reviewing now suggests a simple “analytic” or algebraic proof.
The conic sections of Apollonius are just that, sections of a cone by a plane. As such, they come with diameters; but the cone may be oblique, and then the conjugate diameters may fail to be at right angles to one another. If they are at right angles, they are axes. It is then a theorem of Apollonius that every central conic has axes in this sense. Our exercise uses the idea of the proof.
We are given, or we choose, points (a, b) and (c, d) so that, when we define
δ = ad − bc,
then δ is not 0. We obtain an equation
(dx − cy)2 ± (bx − ay)2 = δ2.
We can understand this to define an ellipse or hyperbola (depending on choice of sign) with conjugate diameters whose endpoints are the original two points and their negatives. In the terminology of Apollonius (concerning which see my article “Abscissas and Ordinates”), those endpoints are vertices. If we translate the center to a third given or chosen point, (e, f), this means replacing x with x − e and y with y − f. We can then rewrite the equation in the general form
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0.
We have now created the exercise of analyzing this equation. By completing squares, we obtain an equation
(αx + βy + κ)2 ± (γy + λ)2 = δ2.
This defines the same conic as before, while displaying a new pair of conjugate diameters: these are segments of the horizontal line given by
γy + λ = 0
and the non-horizontal line given by
αx + βy + κ = 0,
where α ≠ 0. These lines will intersect at (e, f). Translating this back to the origin, we have the conic defined by
(αx + βy)2 ± (γy)2 = δ2.
In fact δ is an invariant of the conic, and the endpoints of the diameters will be (±γ, 0) and (±β, ∓α). In any case, there is a circle defined by the equation
(αx)2 + (αy)2 = δ2.
The circle shares its horizontal diameter with the conic and also intersects the conic in two more points. The four points of intersection are the vertices of a rectangle. One can now compute the slopes of the sides of that rectangle. The axes of the conic have these slopes, and we can now compute their endpoints.
That’s the exercise, in a nutshell. I was honor-bound to work out some examples for myself, before asking the same of students. The computations can be fiendishly difficult to get right. Though having chosen one’s parameters to be single-digit whole numbers, one can end up having to work with square roots of square-free four-digit numbers. That in itself may not be a problem; but one has to get the computations for the radicand right in the first place.
One can do the computations with a calculator, but one still must enter the parameters correctly.
One can write a computer program to do all of the computations, given the initial parameters. If a student can write such a program, that’s great. I did it, in order to check every students’ work. (Most students chose parameters unique to them; I did allow students to work in groups, but many did not take this option.)
I wrote my program in LaTeX, using the ifthen
package in order to have every equation and every coordinate typeset in simplest form. I used the pstricks
and pst-eucl
packages to draw the graphs, and the pgf
package to print decimal approximations of coordinates. Doing all of this—learning how to do it—was a lot of work in itself; however, this work does not seem to duplicate what is done anywhere else, in particular by WolframAlpha.
Thus my exercise is practically immune to cheating. To get it right, you probably have to learn some mathematics, even if you have friends who already know the mathematics.
If you have the money, maybe you can pay some unscrupulous person to learn the mathematics for you and apply it, the way you can pay somebody to complete your writing assignments, even your doctoral dissertation. People who want to be ignorant are able to succeed in being so.
In the exercises of his own textbooks, Michael Spivak is fond of asking for complex computations, as for example of the derivative of the function f given by
f (x) = sin (x/(x − sin (x/(x − sin x)))).
This is the last of a collection (in Calculus, 2nd ed., 1980) of eighteen differentiation exercises that are prefaced by the remark,
It took the author 20 minutes to compute the derivatives for the answer section, and it should not take you much longer. Although rapid calculation is not the goal of mathematics, if you hope to treat theoretical applications of the Chain Rule with aplomb, these concrete applications should be child’s play—mathematicians like to pretend that they can’t even add, but most of them can when the have to.
The author of the present blog post spent countless hours over a number of weeks, engaged in both the theory and practice of the exercise being discussed here. Students should expect to spend hours understanding the exercise and working out their own examples.
The payoff is twofold.
- Students can see for themselves when they have got the right answer.
- The right answer is aesthetically pleasing, if symmetry be a form of beauty, which it classically is (see my essay, “On Commensurability and Symmetry”).
In particular, students can plot the endpoints that they find for the axes and see if they fit the original conic; the conic will then be symmetric about the axes.
If you solve Spivak’s exercise, there’s not a good way to see that you have done it right; also, the answer itself is of no interest. You can compare your answer to Spivak’s or somebody else’s. You can also just review your own work carefully, step by step; or you can do the same work again, to see if you get the same results.
In fact I had to do this a number of times in working out my analytic geometry problem. Even when I knew from graphing that there must be an error in my work, I could have a hard time finding it. Eventually I did find it.
Our graduates may never again use the particular mathematics that we try to teach them. If they learn
- to persist until they have solved a problem,
- to verify their solution, according to standards that they themselves recognize,
this is an ability that will benefit them.
Nonetheless, I have a hedonistic theory of education. Learning is a pleasure, however much it may be accompanied by travail. One then has to learn to accept the travail, or turn it into something else.
If there is any pleasure in solving the differentiation exercise above, I think it is the pleasure of having the power to do what is asked.
If one can analyze an arbitrary second-degree polynomial equation in two variables, in order to obtain (in the non-degenerate cases) a conic section with its axis or axes, this is like knowing how to find flowers in a forest.
In fact my students did not seem to appreciate the visual beauty of the exercise that I assigned them. Of the students who chose to undertake the exercise in the first place, most of them could imitate my own computations, using their own parameters; but not many could reach the end without errors. Of those who could, few confirmed their results with a proper graph, even though I did this in my own examples. (I initially posted my examples as typeset by LaTeX; later I posted handwritten versions, with hand-drawn graphs.)
Some students did plot vertices that obviously did not fit their curves; I wish they could have said (and thus I wish I had encouraged them to say), “I see that I must have made a mistake, but I cannot find it.”
Perhaps I must conclude that my exercise is too hard, even though I see one of its key concepts—the intersection of a circle and a concentric ellipse with oblique axes—in Weeks and Adkins, Second Course in Algebra (1971), which I used in my freshman year of high school.
My own students might better do my exercise if we could all work together face to face. Even then, time spent on the exercise might be better spent learning, for example, that a linear equation in three variables defines not a line, but a plane. I was warned, in fact, that students have seemed not to learn this from their first-year courses.
However, if students have not learned this, they would seem not to have learned to visualize properly. Part of learning linear algebra is learning what not to try to visualize: for example, how there can be four or more mutually orthogonal coordinate axes. Still, I suppose one does visualize how any two of the axes are at right angles to one another.
One should acknowledge here that the visual may be the palpable, perceived with the hands instead of the eyes; the word visual itself, from Latin, is related to idea from Greek, along with wit and wise from Old English.
My own exercise tries to make the point of Descartes: that algebraic equations describe something geometrical, something real.
Perhaps only algebra and computations have been emphasized in our students’ educations and on the national multiple-choice university entrance exam. Students learn how to find answers, but not how to know that they are right answers, except insofar as a teacher tells them that their technique is correct.
The beauty of mathematics is that we can know for ourselves that our answers are correct. However, society may discourage such independence.
Revised August 25, 2020, and again April 17, 2023
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