Question

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)

Answer 1

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.