Which of the following is a point on the plane curved defined by the parametric equations? x=4t y=12t^2+4t-1 a. (4,7) b. (4,207) c.(-2,4) d.(-2,0)

Question

asked Oct 6, 2020 102k views

1 Answer

Tom Dudfield · Answer 1 · 2020-10-11T12:40:17+0000

Answer:

d.(-2,0)

Explanation:

The given parametric equation is:

x=4t

y=12t^2+4t-1

We make t the subject in the first equation;

t=(x)/(4)

We substitute into the second equation to get:

y=12((x)/(4))^2+4((x)/(4))-1

y=(3)/(4)x^2+x-1

When x=4 ,
y=(3)/(4)(4)^2+4-1=15

When x=-2 ,
y=(3)/(4)(-2)^2+-2-1=0

Therefore the point (-2,0) lies on the given parametric curve.