Question

A curve is defined by the equations:
x = 2+2 cos(t)
y = 1-3 sin^2(t)

Find an equation in x and y for this curve.

I know the answer is y=3/4x^2-3x+1
But how do I get there?

Answer 1

x = 2+2 cos(t) --> Solve for cos(t) --> 2cos(t) = x - 2 --> cos(t) = (x-2)/2

y = 1-3 sin^2(t) --> Solve for sin^2(t) --> -3sin^2(t) = y - 1 --> sin^2(t) = (y-1)/(-3)

Answer 2

Recall the trig identity: sin^2(t) + cos^2(t) = 1

Into that, substitute the results from above:

cos(t) = (x-2)/2 and sin^2(t) = (y-1)/(-3)

Simplify and see what you get.

@Loulita

Answer 3

@Directrix

Since cos(t) is not squared, do I have to substitute it as ((x-2)/2)^2?

Answer 4

yes

Answer 5

I agree.

Answer 6

I am having trouble with the simplification.
Here is where I am:

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

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

@Directrix

Answer 7

@Loulita Is this what you are crunching?

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

Answer 8

Yes it is. I may have simplified ((x-2)/2)^2 incorrectly but I'm not sure.

Answer 9

((x-2)/2)^2 is not equal to (x^2-2x+4)/4. Do you see the error?

Answer 10

What is (x - 2)^2 ?

Answer 11

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

cranks out to

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

Answer 12

Oh! I knew I was making a silly mistake somewhere. Thank you!