Question

Which points are on a plane curve described by the following set of parametric equations?
Select all that apply

x= 3t+4 and y= 2t^2

(1,-2)
(1,2)
(1,7)
(2,10)
(7,2)

Answer 1

ANSWER

The points (1,2) and (7,2) lie on the given curve.

EXPLANATION

The given parametric equations are:

$x = 3t + 4$

and

$y = 2 {t}^{2}$

We make t the subject in the first equation to obtain:

$t = \frac{x - 4}{3}$

We substitute this into the second equation to get:

$y =2{(\frac{x - 4}{3} )}^{2}$

When x=1,

$y = 2 {(\frac{1 - 4}{3} )}^{2} = 2$

When x=2

$y =2{(\frac{2- 4}{3} )}^{2} = \frac{8}{9}$

When x=7,

$y =2{(\frac{7 - 4}{3} )}^{2} = 2$

Therefore the points (1,2) and (7,2) lie on the given curve.

Answer 2

The points are on a plane curve described by the following set of parametric equations are:(1,2), (7,2).

What is Parametric equation?

Given:

x= 3t+4 and y= 2t²

Hence:

x=3t +4 = t=(x-4)/3

y=2t² =y=2[(x-4)/3]

y=2[(x-4)/3]²

When x=1

y=2(1-4/3)²

y=2(-3/3)²

y=2(1)

y=2

When x=7

y=2(7-4/3)²

y=2(3/3)²

y=2(1)

y=2

Therefore the points are on a plane curve described by the following set of parametric equations are:(1,2), (7,2).

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