At a Glance - Points on Graphs of Parametric Equations
We can describe trajectories of things with parametric equations.
Before we go forward, we should review parabolas via Super Mario Brothers. When Mario jumps, we can parameterize his trajectory d(x, y) =
To graph this parametric equation, we graph a bunch of individual points. To find the coordinates of a point we take a t value and find the corresponding x and y values.
How do we know if Mario landed safely on the other side of the canyon? We need to know if point (x_{1}, y_{1}) is on the trajectory.
We're going to reverse the graphing process. We'll start with point (x_{1}, y_{1}) and find the t-value that produced it, if there is one.
To determine if a particular point (x, y) is on a parametric graph we need to find a single value of the parameter t that produces the correct x- and y-values at the same time.
Example 1
Determine if the point (5, -5) is on the graph of the parametric equations x(t) = 4t + 1 y(t) = 2t – 7. |
Example 2
Determine if the point (17, 0) is on the graph of the parametric equations x = 4t + 1 y = 2t – 7. |
Example 3
Determine if the point (4, -8) is on the graph of the parametric equations x(t) = t^{2} y(t) = t^{3}. |
Example 4
Determine if the point (4, 3) is on the graph of the parametric equations x(t) = t^{2} |
Exercise 1
- Determine whether each point is on the graph of the parametric equations
x(t) = t^{3}y(t) = t^{4}
- (8, 16)
- (8, -16)
- (-27, 81)
Exercise 2
- Show that the point (5, -6) is on the graph of the parametric equations
x(t) = t^{2} + 1y(t) = 3t
Exercise 3
- Show that the point (11, 9) is not on the graph of the parametric equations
x(t)= t^{2} + 1y(t) = 3t