The Paradox of Translating a Parabola


Sarah Major

The following is based on problem #3 given in Exploration 2, which states:

Examine graphs for the parabola

for different values of a, b, and c. (a, b, c can be any rational numbers).

Fix the values for a and b, vary c. Make at least 5 graphs on the same axes as you vary c.

Try an animation for the same range.

What is happening mathematicatically?

Can you prove this is a translation and that the shape of the parabola does not change?

For the sake of analyzing the mathematics behind these parabolas, I will use the simplest form possible by assigning a=1 and b=1. c will range from -2 to 2. The following is the graphs of those equations using Graphing Calculator 4.0:

From these graphs, we can see that the value for c determines where the y-intercept for the graph is. These locations make it easy to see what is visually happening with the graph. Each time c increases by 1, the y-intercept increases by the same amount. This, in turn, means that all of the points are "shifted up" by the same amount. That is, all of the points on the graphs are translated from (x,y) to (x,y+k). But why does adding or subtracting a certain number of units from the equation of a parabola cause the graph to vertically move that same number of units, and why that translation in particular?

Let's break this problem down mathematically. To make the notation easy, let's consider an equation, f(x). Now, say it has an initial point of (x1,y1), where: y1=f(x1).

Now say we are adding a constant, k (which we will say is positive), to our function so that, once added, we have y=f(x)+k. If we plug in our initial point into this new function, we get y1=f(x1)+k. If we plug in our initial point, we get a different y-value. We get the equation y2=f(x1)+k, or y2=y1+k if we substitute.

After this occurs our initial point, (x1,y1), is translated to (x1,y2), and in this case, y2=y1+k. As we can see, the x-value for the second function has stayed the same, so there is no change in the horizontal value, or x-value, at all. However, the y-value has changed by k units, so it has been moved up k units (since k is positive). If k was negative, then we would have a value subtracted from the original function, so the graph would move downwards. See below for a diagram of the process we just went through:

Therefore, since we have successfully generalized this concept and showed how it mathematically works, we can safely say that adding a value to a function translates the function to points that are that many units above the original points while subtracting from a function translates all of the points down that many units.

From the above, we can see that in the context of our original equations, that each graph is 1 unit higher than the one before it because we are adding 1 unit to the whole function. However, how do we know that the shape of the graph is not changing since we are not able to see the same amount of each graph? This answer comes from finding the area contained within the parabola.

The formula for finding the area contained by a parabola (commonly used to find the area under a parabolic arch) is:

where b is the base and h is the height. The height goes between the vertex of the parabola and the base while the base is made from the intersection of a horizontal line segment intersecting both sides of the parabola. See the following diagram made with Geometer's Sketchpad 4.0:


Why does this formula work? It comes from the construction of a rectangle in this area of a parabola. Let's change the diagram to that of a rectangle:

In this diagram, b times h would be the area of the rectangle. In essence, the formula says that the area in this part of the parabola is 2/3 of the area of the rectangle.

How do we know that this formula works? What other mathematical concept do we use to find the area under a graph? The first thing that should come to mind is the use of integration. However, when we integrate, we are finding the area UNDER the curve in reference to the x-axis. In the case of our graph above, the x-axis constitutes the TOP of our area. This means that if we integrate this particular parabola, we will get a negative area. The absolute value of this value will be the actual area that we are trying to calculate. Let's see what we get when we integrate this particular parabola:



And what would the area be using our formula?


So the formula works. Will we have the same area if we add 4? Be mindful that we must choose one of our parameters, either base or height, to be the same as our first parabola to make the formula useful.


So area is preserved when the parabola is shifted vertically.

Is there another way to see if the shape of a parabola is preserved if we move it vertically? This answer can come from looking at the definition of a parabola. We define a parabola as the set of points that are equidistant from the directrix and the focus. If we know that these distances are preserved when the parabola is shifted vertically, we'll know that the shape is preserved. But how do we find the directrix and focus of our parabola?

Most students are used to working with the vertex form of a parabola equation, which is:

where a is the stretch or shrink of the graph and (h,k) is the vertex. a is also the same a that appears in the general form of the parabola.

A form that students are usually not as familiar with is the conic form:

where, like the other form, (h,k) is the vertex, but p is the vertical distance from the vertex to the focus, making it also the vertical distance from the vertex to the directrix.

Is there a way to find p without having to convert the equation into the conic form? There is a way to find the equivalent of it in either the vertex form or the general form for the parabola:


So, essentially, 1/a = 4p. Therefore, if we want to find p using either the general form or the vertex form, we simply find the reciprocal of the value corresponding to a and divide it by 4.

In the context of the parabolas that we have been using, a = 1, so p = 1/4. We can then find the focus, which will be 1/4 units above the vertex, and the directrix, which will be 1/4 below the vertex. First, we must put the equation into vertex form to find the vertex:

So the vertex is (-0.5,-2.25). From this equation, a=1, so p=0.25. Since p is the distance from the vertex to the focus and also from the vertex to the directrix, we can use this to find the point that is our focus and the line that is our directrix. The focus would be (-0.5,-2) and our directrix would be the line y=-2.50. Let's graph these along with our original parabola:

One can visually see that the vertex of this parabola is equidistant from the vertex and the directrix. We can also pick some random points on the parabola to find the distance from them to our directrix and focus so that when we translate our parabola, we can see if ths distance is preserved and thus confirming that the shape and size is also preserved. Let's pick three points, two on the left side of the parabola and one on the right side. These points will be (-2,0), (-3,4), and (0.5,-1.25):

We first need to calculate the distance from each of the points to both the focus and the directrix. The distance from each point to the focus is easy because we can just use the distance formula:

The distance from the point to the directrix, however, is a bit more complicated because we are not given a certain point on the directrix in which to calculate the distance. We must find the path of least distance from the points to the line that is the directrix. We can do this by using linear algebra.

To do this, we first pick two points on the line, (x1,y1) and (x2,y2). Then we must find the vector perpindicular to the line, which will be given by:

Next we find the vector from the point in question, which we will call for the moment (x0,y0), to the first point on the line. This is given by:

Then, to find the distance from our initial point to the line, we simply project r onto v, which we will call d, by doing the following:


The two points on the line we choose do not matter. They are just specifically so that a vector perpindicular to it can be found. For my calculations, I will call the distance from a point to the focus df and the distance from a point to the directrix dd. Therefore, we get the following distances:

For (-2,0): df = 2.5, dd = 2.5

For (-3,4): df = 6.5, dd = 6.5

For (0.5,-1.25): df = 1.25, dd = 1.25

Just like the definition of a parabola states, the two distances for each point are the same. Therefore, it is not necessary to calculate them both, but for the sake of our investigation, we will to make sure they are the same.

Now, we need to shift the parabola vertically, say four units, and do the same calculations from the corresponding points on the new parabola to its new focus and directrix. Let's first shift our parabola. If we do a vertical translation of four units up, our new equation will be:


And the graph will be as follows:

Let's put our equation in vertex form so that we can find the vertex:


So the vertex is (-0.5,1.75). Since the a value is still the same, p is still 0.25, which is a good sign for what we are trying to show. We can now graph our focus and directrix. The focus will be the point (-0.5,2) and the directrix will be the line y = 1.5:

Now we need to find our corresponding points and make sure they lie on the graph of the parabola. Since the graph was shifted vertically up 4 units, our points should be (-2,4), (-3,8), and (0.5,2.75):

They all do appear on the parabola. Now just to find the distances from each point to the focus and from each point to the directrix:

For (-2,4): df = 2.5, dd = 2.5

For (-3,8): df = 6.5, dd = 6.5

For (0.5,2.75): df = 1.25, dd = 1.25

So, the distances for a translated parabola that has been shifted vertically are the same. Thus, the size and shape of the parabola has been preserved.

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