Level curves of quadratic functions

We’re now going to discuss in some detail what are called conic sections. Why conic sections? Well, the ancient Greeks had found long ago that if you took a cone and if you sliced it by a plane, then depending on the positioning and the angle of the plane with which you did the slicing, you would either get an ellipse or possibly a parabola or maybe a hyperbola– the two branches of a hyperbola. Now, personally, I attach no great importance to this intriguing fact, but the curves themselves are undoubtedly of the greatest importance in a wide range of applications and optics and mechanics and so forth. They are truly fundamental mathematical objects.

Now I have to give great credit to these ancient Greek mathematicians for sensing the importance of these curves and for being able to study them so completely. Let’s summarize some things we basically know about the conic sections. The first class that we looked at was the ellipse. The canonical equation of an ellipse, which is defined in a certain geometrical way that I will not go back on, the canonical equation is of the form x squared over a squared plus y squared over b squared equals 1. As you know, this means that a point, whose coordinates are x, y, lies on the ellipse if and only if the coordinates satisfy this equation.

If the a is greater than b in this equation, then the ellipse is horizontally oriented. If the a is less than b, then the ellipse is vertically oriented. And if a is equal b, the equation turns out to be that of a circle. So in this sense, you can consider a circle to be a special case of an ellipse. Then we had looked at the parabola. The canonical equation of a parabola, initially that we found, was y equals x squared over 4 p. And if p is a positive parameter, that gives us our upward opening parabola that we saw before. The same equation if p is negative, gives you a parabola but one that opens downward instead of upwards.

Now you notice in the equation of a parabola, there’s not perfect symmetry between the y and the x. One is squared and one isn’t. One can easily consider a parabola in which it’s the x that’s equal to y squared over 4 p, for instance. In that case, if p is positive you’ll get a rightward horizontally oriented parabola. And if p is negative, you’ll get a leftward opening parabola. And finally, we had looked at the hyperbola. Its canonical equation has that minus sign in front of the y squared, as you know.

If you now allow the possibility of plus or minus on the right hand side– before we only had plus 1– you cover the two cases, the horizontal and the vertical. That is, if you take the plus 1 as we had before, you get a horizontally oriented hyperbola, which as you know has two branches, 1 leftward and 1 rightward opening. And if you take the minus 1 in the canonical equation, you get a vertically oriented hyperbola.

Now if someone gives you a quadratic equation of this type– that is, an equation involving two variables, x and y– and you see that there are terms with x squared, y squared, also x y in general is a quadratic term, and then some linear terms, d x and e y in a constant and ask you to find the level curve of that quadratic function of two variables. How do we do it? Well, to begin with, we assume that a, b, and c are not all zero, because they define what are called the quadratic or ordered two terms here. And we don’t want them all to be zero.

Otherwise, we’d just have an affine function with only linear terms in x and y. And we know the equation would be a straight line. So we’ll assume a, b, and c are not all zero. And we want to know what kind of curve you get from this equation. Well, if b is equal 0, that’s the coefficient of x y. There are no mixed terms. That means that the only quadratic terms are x squared and y squared really present. Then you can rearrange the equation. Here, of course, completing the square is the basic procedure.

The equation will then look like one of the canonical equations for a conic curve, except that it may be centered at some point different from 0, 0. In each case, there will always be the possibility that no solutions exist. We saw such an example for the circle some time back. That’s a degenerate case, we call it, but most of the time there should be a conic curve. That’s if b is equal zero, a rather simple case. If b is different from zero, then it’s more complicated. But we can state the following fact.

If you construct the determinant– the discriminant, rather– that’s the expression b squared minus 4 a c, just as it was for quadratic functions and equations that we were solving. This discriminant tells you the type of conic curve to expect. Now when b is different from zero, the curve will wind up being skewed or rotated. But it will still be one of the three main types unless, of course, it’s degenerate, as we’ll see. How does the discriminant work here? Well, you calculate b squared minus 4 a c, which is easy to do, and if that turns out to be strictly negative, then you expect an elliptic curve. If it’s exactly zero, a parabola, and if it’s strictly positive, a hyperbolic curve.

Let’s examine each of these three cases briefly to see what happens. The elliptic case, then, is when the discriminant is strictly negative. For example, you can calculate here the discriminant for this quadratic equation. The b here is equal zero. As you see, there’s no x y term, And a and c are both positive. a is 2, and c is 2. So the discriminant is strictly negative. What do you expect when the discriminant is negative? Well, an ellipse, generally speaking. But here you can even be more discerning and expect a circle. Why? Because x squared and y squared have equal coefficients plus 2. So it should be the special case of an ellipse given by a circle.

In fact, when you complete the squares and work it out, I’ll let you check that you’ll get a circle centred at a certain point that I expect you to be able to find and of a certain radius, which again, you should be able to calculate. Now here’s another equation where you can also see that the discriminant is negative. In this case, the b here is equal to minus 2. And the discriminant turns out negative. Therefore, you expect an ellipse, and it shouldn’t be a circle because x squared and y squared have different coefficients. And, in fact, you do get an ellipse.

If you use your favourite graph application to draw graphs, it’ll tell you that you do get an ellipse, but it’s skewed. You notice it’s neither horizontal nor vertically oriented. That’s because of the term minus 2 x y in the quadratic equation. Here’s another example. And you can see that this looks right and possibly looks like an ellipse. And you can see that the discriminant is negative. However, it’s an impossible equation. The sum of 3 positive numbers can’t give you zero, especially with the 7 there. So this is the degenerate case that arises in the elliptic case. Degenerate case means what? It means that you’ll get an empty graph. There are no points x y that satisfy your equation.

And so you don’t actually see an ellipse. You see nothing, the empty set. Now let’s look at the parabolic case. The parabolic case corresponds to the discriminant b squared minus 4 a c being exactly zero. Here’s an example. I’ll let you check. Minus 4 squared minus 4 times 4 times 1. Yes, you get zero. The discriminant is zero in this case. So you should expect what? A parabola. In fact, when you graph this thing, you wind up getting a parabola, but it’s skewed, as you can see. It’s neither horizontal nor vertical. That’s because of the x y term in the equation.

Here’s another equation in which you would expect a parabola to occur, because as you will check, the discriminant turns out to be zero in this equation. But it’s actually going to be degenerate for the following reason. It turns out that the left hand side of this equation can be factored exactly into the product of two affine expressions in x y.

And you notice that 2 x plus y is present in each one. So when is this going to be zero? It’s going to be zero when either of these two factors is zero. But the first factor being zero, for example, just corresponds to x y being on a certain line, because it’s an affine equation. The same thing for the second, and you see that the lines have the same slope. Therefore, they’re parallel. So in other words, this equation holds if and only if you’re on either of two parallel lines. So the solution set is the union of those two lines. And this is how the parabolic case can be degenerate.

It can lead to the union of two parallel lines such as, in this case, this set.

And finally, there’s the hyperbolic case to consider. That’s the case in which the discriminant is strictly positive. And here’s an example. You calculate the discriminant. You’ll see it’s strictly positive , and you expect, therefore, a hyperbola. That can be graphed, and it confirms your suspicion. So it’s rotated somewhat, the hyperbola, because of the x y term. Here’s another example. This turns out to be degenerate in this case, why? Because again it factors into two affine expressions, but this time the two affine expressions correspond to lines that are not parallel. To say that this product is zero, therefore, is to say either that you’re on the first line or that you’re on the second non-parallel line.

That is, that you’re somewhere on the union of two non-parallel lines. That’s how the hyperbolic case can be degenerate. The solution set can be the union of two intersecting, since they’re not parallel, lines. Now the full proof of these facts is fairly advanced. You might meet it at university in what is called a linear algebra and matrices course on those subjects. But even knowing these facts, I assure you, puts you in the true conic elite.