Points on Graphs of Parametric Equations Examples

Determine if the point (5, -5) is on the graph of the parametric equations

 x(t) = 4t + 1 

y(t) = 2t – 7.

If x = 5 then 5 = 4t + 1 therefore t must equal 1. 

Check to see if t = 1 produces the the correct y-value. 

When t = 1, y = 2(1) – 7 = -5. This is the correct y-value. 

We conclude that the point (5, -5) is on the graph and occurs when t = 1.

Determine if the point (17, 0) is on the graph of the parametric equations x = 4t + 1 y = 2t – 7.

If x = 17 then

17 = 4t + 1

therefore t = 4.

Let's see what we find from the y equation when t = 4. We find

y = 2(4) – 7 = 1.

Since t = 4 is the only value of the parameter that makes x = 17, and at this time y ≠ 0, the point (17, 0) is not on the graph. There's no value of t that makes x = 17 and y = 0 at the same time.

When the parametric equations aren't linear, there may be more than one value of t that could generate the point in question.

Determine if the point (4, -8) is on the graph of the parametric equations 

x(t) = t²

y(t) = t³.

When x = 4 we can have either t = 2 or t = -2. We need to see if either of these t-values produces y = -8.

When t = 2 we find

y = (2)³ = 8,

which isn't what we want. When t = -2 we find

y = (-2)³ = -8,

which is what we want. The point (4, -8) is on the graph of the parametric equations and occurs when t = -2.

If the equation for y is nicer, or has few solutions, than the equation for x, we might use the y-equation first to find the value of t.

Determine if the point (4, 3) is on the graph of the parametric equations

x(t) = t²
y(t) = t + 2.

If we solve the x equation for t we'll find two possible t-values to check. If we solve the y equation we only find one. When y = 3 we have t = 1. Then x = (1)² = 1. Since the only value of t that makes y = 3 doesn't make x = 4, the point (4, 3) isn't on the graph.