Asked by Aditya
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?
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?
Answers
Answered by
Steve
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
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
Answered by
Reiny
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
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
Answered by
Aditya
For both theta = pi/2 and 3pi/2, r = 2. Does this mean that its only half an ellipse?
Answered by
Damon
.... and how did you make out with the spherical coordinates problem Aditya?
Answered by
Damon
and no it is the whole ellipse