The graph of 4/(2-cos(theta)) forms a closed curve. The area of the region inside the curve can be expressed in the form k \pi. What is k^2?

So I tried this problem and I got it to be an ellipse 1 = (((x + 8/6)^2)/((16/3)^2)) + ((y^2)/4)) Is this correct?

Hmm.

when θ=0, r = 4
when θ=pi, r = 4/3

so, the major axis is 16/3, and the semi-major axis is 8/3, with center at x=4/3, so it will be

(x-4/3)^2/(8/3)^2 + y^2/b^2 = 1

and you can figure out b.

As for the area, for an ellipse with semi-major axes a and b, area = pi*a*b

First of all I will assume that your equation is in polar form, so it should have been

r = 4/(2 - cosØ)

converting:
r = 4/(2 - x/r)
2r - x = 4
2r = x+4
square both sides
4r^2 = x^2 + 8x + 16
4(x^2 + y^2) = x^2 + 8x + 16
3x^2 - 8x + 4y^2 = 16
3(x^2 - (8/3)x + 16/9) + 4y^2 = 16 + 16/3
3(x - 4/3)^2 + y + 4y^2 = 64/3
(x-4/3)^2/(64/9 + y^2/(16/3) = 1

(x-4)^2 / (8/3)^2 + y^2 / (4/√3)^2 = 1

you were close, yes it is an ellipse
http://www.wolframalpha.com/input/?i=plot+%28x-4%2F3%29%5E2%2F%288%2F3%29%5E2+%2B+y%5E2%2F%284%2F√3%29%5E2+%3D+1

so a = 8/3 and b = 4/√3

area of ellipse = abπ
= (8/3)(4/√3)π
= (32/(3√3))π

now you want this in the form k/π
so multiply my answer by π/π to get a π at the bottom

leave it up to you to finish

also a good idea to check all my arithmetic, since we had a slightly different answer to our ellipse

For both theta = pi/2 and 3pi/2, r = 2. Does this mean that its only half an ellipse?

.... and how did you make out with the spherical coordinates problem Aditya?

and no it is the whole ellipse

To determine if your answer is correct, let's start by simplifying the equation of the graph:

Given: 4/(2 - cos(theta))

To simplify, we can multiply both the numerator and denominator by (2 + cos(theta)):

4/(2 - cos(theta)) * (2 + cos(theta))/(2 + cos(theta))

This simplifies to:

8 + 4cos(theta) / (4 - cos^2(theta))

Next, we can apply the Pythagorean identity cos^2(theta) = 1 - sin^2(theta):

8 + 4cos(theta) / (4 - (1 - sin^2(theta)))

Simplifying further:

8 + 4cos(theta) / (3 + sin^2(theta))

Now we have an expression involving both cos(theta) and sin(theta). To convert this into an equation involving x and y, we can use the trigonometric identities:

cos(theta) = x / r
sin(theta) = y / r

Where r is the radius or magnitude of the vector (x, y). We can replace cos(theta) and sin(theta) in our expression:

8 + 4(x / r) / (3 + (y / r)^2)

To eliminate the r term, we can square both sides of the equations:

x^2 + y^2 = r^2

Now we can express r^2 in terms of x and y:

r^2 = x^2 + y^2

Substituting this back into the expression:

8 + 4(x / sqrt(x^2 + y^2)) / (3 + (y / sqrt(x^2 + y^2))^2)

Simplifying further:

8 + 4x / sqrt(x^2 + y^2) / (3 + y^2 / (x^2 + y^2))

After some more algebraic manipulation, the equation becomes:

8(x^2 + y^2) + 4x(x^2 + y^2) / (3(x^2 + y^2) + y^2)

Simplifying further:

(8x^2 + 8y^2 + 4x^3 + 4xy^2) / (3x^2 + 4y^2)

Comparing this to the general formula for an ellipse:

((x + h)^2 / a^2) + (y^2 / b^2) = 1

We can see that the equation is not in the standard form of an ellipse. Therefore, your answer is not correct.

To find the correct equation for the graph, we would need to go back to the original expression and perform further algebraic manipulation. However, determining the value of k^2 based on the area of the region inside the curve is impossible without additional information or calculations.