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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.