Department of Mathematics Education
J. Wilson, EMAT 6680

An Introduction to Parametric Equations
by David Wise

This investigation serves as an introductory lesson in parametric equations. Given a point and slope, students that have successfully completed Algebra I, should be able to easily write the equation of the line. The two goals of this investigation are for students to learn how to write parametric equations given a point and slope, and for students to gain an ability to interpret parametric equations.

These demonstration graphs have been produced using the graphing calculator software, Graphing Calculator 2.2.1. If you do not have this software, you can use a demo version, or purchase the software through the internet at These demonstrations can also be created using other graphing software, or graphing calulators.

Begin by first writing the basic linear equation. y = mx + b

Place the slope, x-coordinate, and y-coordinate into the equation and solve for b. 5 = 3(7) + b, so b = -16

Therefore, the linear equation is: y = 3x - 16.

Now, we need to manipulate the linear equation to develop the corresponding parametric equations. In parametric equations, the x-variable and y-variable are each written as a function of a third variable, t, called the parameter. Parametric equations allow us to determine a specific x-cooridinate and y-coordinate of a point based on a specific t value. In many applications, t represents time, which can be a helpful way to think of t. Using this concept, parametric equations allow us to determine the location of a specific point (x, y) at a specific time (t).

We need to introduce the variable t.

One way is to simply set t equal to x. The resulting parametric equations are:
x = t
y = 3t - 16

What do these equations mean and are they really the same as the linear equation that we have grown so comfortable in dealing with?

Let's construct a table of values.

t-value x-coordinate y-coordinate
0 0 -16
1 1 -13
2 2 -10
3 3 -7
4 4 -4
5 5 -1
6 6 2
7 7 5

Notice that we obtain the same x- and y-coordinates from the linear equation and the parametric equations. The key difference is that in the parametric equations, the x- and y-coordinates are associated with a t-value. We can think about each point in terms of "time". For example, at time 0, we get the point (0, 10) and at time 7, we get the point (7, 5).

The major difference in graphing a linear equation and the corresponding parametric equations is the restriction we place on t. When we graph a linear equation, we obtain a line because the domain and range consists of all real numbers. However, with parametric equations, t is usually restricted, so that the graph produced focuses on the specified values of t. For example, Graphing Calculator 2.2.1 has a default, so that t min = 0 and t max = 1. Therefore, instead of obtaining the line y = 3x - 16, we obtain a line segment of y = 3x - 16. The only way to obtain the line y = 3x - 16 using the parametric equations we produced is to set t min = -infinity and t max = +infinity. Let's look at the graphs.

1. y = 3x - 16

2. x = t and y = 3t - 16 for t min = 0, t max = 1


3. x = t and y = 3t - 16 for t min = 0, t max = 7


In the third set of graphs, the endpoint (7, 5) of the line segment corresponds with t = 7. The preceeding graphs should have convinced us that the linear equation and parametric equations produce produce the same line when t is infinite, but if not try the following two tricks. First, if we try to graph the linear equation and the corresponding parametric equations on the same coordinate plane, the graph of the parametric equations is "hidden" by the linear equation. Second, when graphing the parametric equations, set t min equal to a "very large" negative number and t max equal to a very large positive number.

Note: If you are a beginning user of Graphing Calculator 2.2.1, in order to enter parametric equations, you must use the 2-Vector function of the Equation menu. Also, in order to change t, you must use the View menu and Parameters.

No, without too much difficulty, we can develop another set of parametric equations that satisfies the given conditions.

We know that the linear equation is: y = 3x - 16.

The key is to consider factoring. y = 3x - 15 - 1, so y = 3(x - 5) - 1.

We now can introduce the variable t. In the previous example we set t = x, this time we will set t = x - 5. The resulting parametric equations are:
x = t + 5 -----(from, t = x - 5)
y = 3t - 1

Let's construct a table of values to verify that this set of parametric equations correspond with the linear equation and previous set of parametric equations. The trick is to try to find the t-values that correspond to the x-coordinates we found previously.

t-value x-coordinate y-coordinate
-5 0 -16
-4 1 -13
-3 2 -10
-2 3 -7
-1 4 -4
0 5 -1
1 6 2
2 7 5

Notice that we obtain the exact same points as we did in the previous example. The only thing that has changed is the t-value. Therefore, this set of parametric equations produces the same line, but we obtain the same points at different times. We have changed what the parameter (t) is equal to, meaning that we have changed the parameter. Consider the point (7, 5), which was the point that our line had to pass through. With the first set of parametric equations, we obtained the point (7, 5) at time t = 7. With our new set of parametric equations, we obtained the point (7, 5) at time t = 2.

We can also varify this graphically.

x = t + 5 and y = 3t - 1 for t min = -5, t max = 2



  1. How many possible sets of parametric equations can be created based upon a linear equation? Explain.
  2. Consider the two parametric equations developed in this investigation. Try to establish a connection between the coefficients and constants within each set of parametric equations to the graph produced. Hint: Restrict the t-value to t-min = 0 and t-max = 1. You should be able to find a general rule between the set of parametric equations and generating its graph. Meaning that k, a, and b, describe particular important aspects of the graph of the parametric equations, and therefore, provide the ability to "visualize" the graph without actually having to graph them.

x = t
y = 3t - 16

x = t + 5
y = 3t - 1

x = t + a
y = kt + b

If you have any comments concerning this investigation that would be useful, especially for use at the high school level, please send e-mail to

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