To find out if a point is on the plane curve defined by the parametric equations, we need to substitute the given values of x and y into the parametric equations and see if they are satisfied.
For the point (4,7):
x = 4
y = 7
Substitute into the parametric equations:
4 = 4t
7 = 12t^2 + 4t - 1
From x = 4 = 4t, we get t = 1.
Substitute t = 1 into y = 12t^2 + 4t - 1:
y = 12(1)^2 + 4(1) - 1 = 12 + 4 - 1 = 15
Therefore, the point (4,7) is not on the plane curve defined by the parametric equations.
For the point (4,207):
x = 4
y = 207
Substitute into the parametric equations:
4 = 4t
207 = 12t^2 + 4t - 1
From x = 4 = 4t, we get t = 1.
Substitute t = 1 into y = 12t^2 + 4t - 1:
y = 12(1)^2 + 4(1) - 1 = 12 + 4 - 1 = 15
Therefore, the point (4,207) is not on the plane curve defined by the parametric equations.
For the point (-2,4):
x = -2
y = 4
Substitute into the parametric equations:
-2 = 4t
4 = 12t^2 + 4t - 1
From x = -2 = 4t, we get t = -0.5.
Substitute t = -0.5 into y = 12t^2 + 4t - 1:
y = 12(-0.5)^2 + 4(-0.5) - 1 = 3 - 2 - 1 = 0
Therefore, the point (-2,4) is on the plane curve defined by the parametric equations.
For the point (-2,0):
x = -2
y = 0
Substitute into the parametric equations:
-2 = 4t
0 = 12t^2 + 4t - 1
From x = -2 = 4t, we get t = -0.5.
Substitute t = -0.5 into y = 12t^2 + 4t - 1:
y = 12(-0.5)^2 + 4(-0.5) - 1 = 3 - 2 - 1 = 0
Therefore, the point (-2,0) is also on the plane curve defined by the parametric equations.
Therefore, the points (-2,4) and (-2,0) are points on the plane curve defined by the parametric equations.
which of the following is a point on the plane curve defined by the parametric equations?
x=4t
y=12t^2+4t-1
*(4,7)
*(4,207)
*(-2,4)
*(-2,0)
1 answer
1 answer