Question

Which of the following is a point on the plane curved defined by the parametric equations?

Answer 1

Answer:

d.(-2,0)

Step-by-step explanation:

The given parametric equation is:

$x=4t$

$y=12t^2+4t-1$

We make t the subject in the first equation;

$t=\frac{x}{4}$

We substitute into the second equation to get:

$y=12(\frac{x}{4})^2+4(\frac{x}{4})-1$

$y=\frac{3}{4}x^2+x-1$

When x=4 , $y=\frac{3}{4}(4)^2+4-1=15$

When x=-2 , $y=\frac{3}{4}(-2)^2+-2-1=0$

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