A curve is described by the equations {x = 2 + 2 cos t and y = 1 – 3 sin^2 t} with 0 ≤ t < 2π}.

Find an equation in x and y for this curve.

The solution that the book gives y = 3/4x^2 - 3x + 1.

I know that the graph gives a section of a parabola, I would like to know how they got that answer and the steps that it took to answer it. A thorough explanation is greatly appreciated.

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The given curve is {x = 2 + 2 cos t and y = 1 – 3 sin^2 t} with 0 ≤ t < 2π}.

We have to find the equation of the curve in terms of x and y.

To do this, we eliminate t. For this , we find cost (=cosine of t) from the first equation and then we find (cost)^2. We find (sint)^2 from the second equation. Then we use (cost)^2+(sint)^2 = 1 to eliminate t.

From the equation x = 2+2cost, we get x-2 = 2cost. Therefore solving for cost we get:

cost = (x-2)/2....(1).

From the equation y = 1-3(sint)^2, we have 3(sint)^2 = 1-2. From this we solve for (sint)^2 :

(sint)^2 = (1-y)/3.........(2)

From (1) we get:

(cost)^2 = {(x-2)/2}^2 .....(3).

Adding (2) and (3) we get:

(sint)^2+ (cost)^2 = (1-y)/3 +{(x-2)/2)}^2.

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

12 = 4(1-y) + 3(x-2)^2

4y-4+12 = 3(x-2) ^2

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

4(y+2) = 3x^2-12x+12.

We divide by 4:

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

We subtract 2:

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

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

Hope this helps.

We have the equations for x and y as x = 2 + 2 cos t and y = 1 – 3 sin^2 t.

Now we have to eliminate t and form an equation with only x and y.

Take x = 2 + 2 cos t

=> x- 2 = 2 cos t

raise both the sides to the power 2

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

Open the brackets and convert cos t to sin t, this can be done using the relation (cos t)^2 = 1 - (sin t)^2

=> x^2 + 4 - 4x = 4* [ 1 - (sin t)^2]

=> x^2 + 4 - 4x = 4 - 4*(sin t)^2

=> x^2 /4 - x = -(sin t)^2

=> (sin t)^2 = x - x^2 / 4

Now take y = 1 – 3 sin^2 t

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

=> 3* (sin t) ^2 = 1 - y

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

Equate the two (sin t) ^2

=>1/3 - y/3 = x - x^2 / 4

=> - y/3 = x - x^2 / 4 -1/3

=> y/3 = x^2 / 4 - x + 1/3

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

**Therefore we get the result in terms of x and y as y = (3/4)*x^2 - 3x + 1**

We have to mention that the given equations are called parametric equations:

x = f(t) and y = g(t), where t is the parameter.

x = 2 + 2 cos t (1)

y = 1 – 3 sin^2 t (2)

To determine the equation of the line, we'll have to eliminate the parameter t.

Since the equation for y contains the term 3 sin^2 t, we'll try to obtain the same term, but with opposite sign, in the equation for x.

The first step will be to isolate the term in t to the right side and to square raise, both sides, the equation (1):

(x-2)^2 = (2 cos t)^2

We'll expand the square from the right side:

x^2 - 4x + 4 = 4 (cos t)^2 (3)

We'll isolate 3(sin t)^2 to the right side, in equation (2):

1 - y = 3(sin t)^2 (4)

We'll multiply (3) by 3 and (4) by 4:

3x^2 - 12x + 12 = 12 (cos t)^2 (5)

4 - 4y = 12(sin t)^2 (6)

We'll add (5)+(6):

3x^2 - 12x + 12+4 - 4y=12 (cos t)^2+12(sin t)^2

We'll factorize by 12 to the right side:

3x^2 - 12x + 12+4 - 4y=12[(sin t)^2 + (cos t)^2]

From the fundamental formula of trigonometry, we'll have:

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

3x^2 - 12x + 12+4 - 4y=12

We'll subtract 12 both sides:

3x^2 - 12x + 12 + 4 - 4y - 12 = 0

We'll combine and eliminate like terms:

3x^2 - 12x - 4y + 4 = 0

We'll add 4y both sides and we'll use symmetric property:

4y = 3x^2 - 12x + 4

We'll divide by 4:

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

**The equation of the line described by the parametric equations, x = f(t) and y = g(t), is:**

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

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