The parabola is one of four conic sections that I intend to provide a discourse on. The conic sections are so named because they can be obtained by passing a plane at varying angles through a double-napped right circular conical surface. The intersection of the plane and the surface is the conic section. In the case of the parabola, the plane passes through one half of the surface at an angle such that the plane is parallel to a generating straight line on the conical surface as seen below.
In the real world parabolas are typically seen when any projectile object travels through the air and is subjected to the pull of gravity (if air resistance is negligible). If you have ever seen a long exposure photograph of a bouncing ball that is moving laterally, you have seen a series of parabolas. Or if you have observed a water fountain flow as the water goes up and then down due to gravity. Also, the golden McDonald’s arches are a pair of parabolas. Paraboloids, the three- dimensional cousin of the parabola, is a common shape used in satellite dishes and headlight beam reflective surfaces. In highway engineering, the parabola is used to design vertical curves or hillcrests.
We can derive an algebraic expression for the parabola using the definition that a parabola is the set of all ordered pairs or points that lie in a plane that are equidistant from a fixed point called the focus and a fixed line called the directrix that also lie in the plane. (Provided the focus does not lie on the directrix) A line in the plane that is perpendicular or orthogonal to the directrix that passes through the focus is called the axis of symmetry or simply the axis. In layman’s terms it divides the parabola into two halves that are mirror images of one another. The point on the parabola that the axis intersects is called the vertex.
Consider a parabola whose vertex is at the origin (0,0) and has a focus at the point (0,f). Intuitively then we know that the distance from the vertex to the focus is f. (This is also called the focal distance.) We also see that this parabola will open along the y-axis. Since the vertex lies on the parabola it must be equidistant from the focus and directrix. Consequently we know that the directrix must be the line y = -f. Next using the distance formula, we can say the distance from a random point in the plane (x,y) to the focus (0,f) is given by:
Next realize that the point on our directrix that falls vertically below the point (x,y) is given by (x,-f). Then using the same methodology as above we can say that the vertical distance from the directrix to our point (x,y) is given by:
By the definition of a parabola we can set these two values equal to one another and resolve to standard form:
Through similar reasoning we could derive the expression for a parabola with a vertex at the origin that opens along the x-axis using a focus point of (f,0) and a directrix of x = -f. As might be expected this will yield:
Next let us consider how we can develop expressions for a parabola with a vertex that is not at the origin. The methodology utilized to consider a vertex at any general point in the plane is usually called translation of axes.
Translation of Axes
Consider the above two axes. The base x and y axes are shifted to a new general location where the new origin, O’, is at the base coordinate position (h,k). The point P can be described in both coordinate systems and the two systems can be related as follows:
Using these short equations in a substitutionary fashion we can go through the same procedure as above and derive an expression for a parabola with a vertical axis with a vertex at O’ or base coordinate (h,k). This yields:
Likewise, for a parabola with a horizontal axis we obtain:
Rotation of Axes
The other transformation we can do is called rotation of axes and it allows us to generate an expression for a parabola with an axis at an angle other than vertical or horizontal.
As with translation we want to develop a relationship between the two coordinate systems as they describe the same point in the plane. Observing the diagram above we will define the length of the blue line from the origin to the point P as having the magnitude d. Next consider the triangle OPQ’. Using this right triangle we can say:
Then look at triangle OPQ. Similarly as above we can say:
Using the trigonometry identities:
We can write an expressions for x and y and substitute using x’ and y’ to obtain expressions of the relationship between the two systems.
Finally using these last two expressions we will solve for x’ and y’. This will allow us to generate a parabola on the X’ and Y’ axes and then make a substitution in order to find the expression for the parabola in the base coordinate system. The derivation is done by solving each of the last two expressions for x’ and y’ respectively. This will result in four equations:
Next we will set x’ from the one equation equal to x’ from the other and solve for y’. Likewise we will set y’ equal to y’ and then solve for x’.
The last step in each of the above derivations is accomplished by the following identity:
In a similar fashion as above with translation we can substitute the two boxed equations into the general parabola equation to derive a general equation for a rotated parabola. An important caveat to remember with regard to this formula is that the angle can only be an acute angle. However this does not pose a problem since any parabola in any quadrant can be obtained by an acute rotation from an adjacent axis. I will not here go through each step to derive the general formula as it is rather lengthy and cumbersome for this format. However here is the end result in implicit form for a parabola whose original axis was vertical:
I hope that this has been helpful and I hope the next time you are using the water fountain you notice the lowly parabola at work.
I would love to hear your questions or comments.