PLEASE HELP
8.04b

1. Eliminate the parameter.
x = 5t, y = t + 8

A) y = 5x + 8
B) y = x divided by five + 8
C) y = 5x - 8
D) y = x divided by five - 8

2. Eliminate the parameter.
x = square root of x , y = 3t + 7

A) y = 3x2 + 7
B) y = 3 square root of x + 7, x ≥0
C) y = 3 square root of x - 7, x ≥0
D) y = 3x2 - 7

3. Eliminate the parameter.
x = t2 + 2, y = t2 - 4

A) y = x - 6, x ≥ 1
B) y = x + 6, x ≥ 1
C) y = x2 - 6, x ≥1
D) y = x2 + 6, x ≥1

4. Eliminate the parameter.
x = 4 cos t, y = 4 sin t

Respuesta :

These are four questions and four answers:

Question 1. Eliminate the parameter. x = 5t, y = t + 8
 
Answer: option B: y = (x / 5) + 8 ↔ x divided by 5 + 8

Explanation:

1) From x = 5t => t = x / 5

2) Substitute t in y = t + 8 => y = (x / 5) + 8

Which is the option B) y = the division of x by 5, added to 8.

Question 2. Eliminate the parameter.

There is an error in the question, because it says x = √x

If the righ question is x = √t , y = 3t + 7 , then the a
nswer is the option A) y = 3x^2 + 7

Explanation:

1) Given: x = √t , y = 3t + 7

2) from x = √t => t = x^2

3) Substitute t = x^2 in y = 3t + 7

=> y = 3 x^2 + 7 ↔ option A) y = x^2 + 7

Question 3. Eliminate the parameter.

x = t^2 + 2, y = t^2 - 4

Answer: option A) y = x - 6

Explanation:

1) From x = t^2 + 2 => t^2 = x - 2

2) Substitute t^2 in y = t^2 - 4

=> y = x - 2 - 4

=> y = x - 6, since t^2  is ≥ 0, then x ≥ 0

=> option A) y = x - 6, x ≥ 1

Question 4. Eliminate the parameter.
x = 4 cos t, y = 4 sin t

Answer: x^2 + y^2 = 16

Explanation:

1) Square both sides of x = 4 cost

=> x^2 = (4 cos t)^2

x^2 = 16 (cos t)^2

2) Square both sides of y = 4 sin t

y^2 = 16 (sin t)^2

3) Add x^2 and  y^2

x^2 + y^2 = 16(cos t)^2 + 16 (sin t)^2

=> x^2 + y^2 = 16 [ (cos t)^2 + (sin t)^2 ]

4) since (cos t)^2 + (sin t)^2 = 1

x^2 + y^2 = 16