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?

Question

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?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

and no it is the whole ellipse