Lines and Parabolas

by Mark Freitag

An interesting question to pose to advanced mathematics high school students is to find two linear functions f(x) and g(x) such that their product h(x) = f(x) g(x) is tangent to each of f(x) and g(x) at two distinct points. This problem may prove to be difficult to students using pencil and paper, but with the program Algebra Xpressor an investigation into this problem is much easier.
To begin to solve this problem, the students may start by experimenting with parallel lines. It should be obvious that vertical or horizontal parallel lines will not work since their product does not result in a parabola. Generally, students should notice that parallel lines will not work because, intuitively, a parabola will get increasingly wider as the points get further away from the directrix. Since parallel lines will always be equal distance from each other, it can be seen that at most one can be tangent to the parabola. If the students can not see the reasons for this situation, they may graph several examples like the following ones.

y = -2x + 1
y = -2x- 3
y = (-2x + 1)(-2x - 3)

y = 3x + 2
y = 3x + 5
y = (3x + 2) (3x + 5)

Since parallel lines do not satisfy the desired conditions, the students must conclude that the lines intersect at some point. They may try several different graphs using the program, and they should look similar to the following.

y = -2x + 1
y = 3x - 3
y = (-2x +1)(3x -3)

y = -4x +2
y = -3x +3
y = (-4x + 2)(-3x +3)

Again, the students will see that, generally, these do not satisfy the conditions of the problem. Hopefully, the next step that the students will take is to conclude something about the symmetry of the parabolas that they have just graphed. It can be seen that the parabolas in all of the previous examples have had a directrix that has been parallel to the to the x-axis. So the parabola is symmetric around a line perpendicular to the x-axis and which passes through its focal point. This leads to the idea that two lines that are symmetrical around a line parallel to the y-axis may be part of the solution to the problem. Lines that would satisfy this condition would be lines that have slopes that are opposite of each other. Again, students can use the computer to graph some of these lines quickly and without error. Some possible examples may look like the following.

y = -4x + 2
y = 4x + 3
y = (-4x + 2)( 4x + 3)

y = -2x - 2
y = 2x + 5
y = (-2x - 2)(2x + 5)

Again, students should notice that while this is closer to the desired situation it is still not correct. By looking at the generated graphs, they should see that the parabola is on the same axis of symmetry as the lines, but it is either too high or too low. So the students might next try to shift the lines up or down. Since the parabola is the product of the two lines, it follows that if one line is shifted up or down, then the resulting parabola will be adjusted accordingly. Again, several trials might be done on the computer before one that looks like it satisfies the desired conditions is found.

y = -2x -2
y = 2x + 4
y = (-2x -2)(2x +4)

y = -2x -1
y = 2x + 3
y = (-2x - 1)(2x +3)

y = -2x - 2
y = 2x + 3
y = (-2x - 2)(2x +3)

While the last graph looks like it has the desired properties, the students should verify that it actually does have them. It is a simple process to verify this algebraically, and the following is an outline of this process. The last graph and its equations will be used as an example. First, the coordinates of the two potential tangent points must be found, which can be done using two separate systems of equations.

y = -2x - 2
y = (-2x - 2)(2x +3) = -4x2 - 10x - 6

y = 2x + 3
y = (-2x -2)(2x + 3)

Let us look at the first system of equations. If the graphs intersect, then they meet at y = -2x -2, for some x. So, -2x - 2 = (-2x - 2)(2x +3), for some x, which must mean that 2x+3 = 1. Solving this equation for x, we find x = -1. Thus, for the first set of equations we have (x, y) = (-1, 0) as the point of intersection for these two equations. In a similar manner, we find that (x, y) = (-3/2, 0) is the point of intersection for the second system of equations.

Second, it must be verified that the lines are tangent to the parabola at their respective points. Calculus is used to solve this part of the problem. Simply take the derivative of the equation for the parabola. So if f(x) = (-2x - 2)(2x + 3) = -4x2 - 10x -6, then the derivative is f '(x) = -8x - 10. Since the derivative is the slope of the tangent line of a curve at a given point, all that needs to be done is to substitute the two x values into the derivative and see if the resulting answers match the slopes of the given lines. So, we take x = -1, and substitute it into the derivative. f '(-1) = -8(-1) -10 = -2. This matches the slope of the corresponding line, and so we see that the line is indeed tangent to the parabola at the point (-1, 0). Similarly, we have that the line y = 2x + 3 is tangent to the parabola at the point (-3/2, 0). This should prove to the student that these lines do indeed satisfy the desired criterion.

A harder question for the students is to generalize this situation. In other words, if given a line that is not horizontal or vertical, can a second line be found such that the product of the two lines is a parabola and both lines are tangent to the parabola in two distinct points. The following is an outline of the algebraic method used to find a generalization.

Let f(x) = ax + b, g(x) = cx + d, and h(x) = f(x)g(x) = (ax + b)(cx + d). Suppose f(x) and h(x) intersect at some point (x1, y1), then f(x1) = h(x1) and g(x1) = cx1 +d = 1. Similarly, suppose that g(x) and h(x) intersect at some point (x2, y2), then g(x2) = h(x2) and so f(x2) = ax2 + b = 1. We use two different systems of equations, but our demonstration will only consider the first.

f(x) = ax +b
h(x) = f(x) g(x) = acx2 + (ad + bc)x + bd

Now, since we want the line to be tangent at the point (x1, y1), we set the derivative h '(x1) = a. We also know that x1 = (1 - d)/ c. So using substitution, we have the equation

a = 2ac ((1 - d)/c) + ad + bc

Some simplification gives the equation, 0 = a(1 - d) + bc. In a similar manner, we can manipulate the other system of equations to give us the equation 0 = c(1 - b) + ad. We can then set these two equations equal to each other and simplify. This results in the following equality, c = -a. Using this fact, the equation 0 = c(1 - b) + ab, and some more algebra we get the equation b + d = 1.
So to generalize, if an equation y = ax + b is given, and we want to find another line such that these two lines are tangent to the parabola resulting from their product, then the line y = -ax + d, where b + d = 1 is the appropriate line.

If the students can get to this point in the analysis of the problem, they should try several systems of equations and visually see that these conditions on a, b, c, and d satisfy the demands of the problem. This investigation should work well with advanced high school mathematics students. It should help the students understand what a derivative of a line really is, and how it can be used to solve investigations into the nature of mathematics. Certainly, there are other ways to investigate mathematics using this situation. Why do the lines and the parabola always intersect at the x-axis? Can we find equations that will satisfy these conditions, but have the parabola opening in a direction other than up or down? The teacher should be able to come up with several other puzzling questions with only a minimum of investigation. This example of the use of the Algebra Xpressor is just one in a myriad of possible uses that will help high school students gain a deeper understanding of mathematics.

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