# EquationsA curve is described by the equations x=2+2 cos t and y=1–3 sin^2t} with 0≤t<2π. Find an equation in x and y for this curve.

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### 1 Answer

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