- 192

- 0

x = a sin x

y = b cos x, a>b>0 is

[tex] L = 4a \int_{1}^{pi/2} \sqrt{1-e^2sin^2x} dx

[/tex]

where e is the eccentricity of the ellipse (e = c/a, where c = [tex]\sqrt{a^2-b^2}[/tex]

I've tried a whole bunch of algebraic and trignometric manipulation but I'm getting the feeling I'm overlooking something. How would you approach this problem? I've tried working backwards and eventually got something of the form:

[tex] L = \int_{1}^{pi/2} \sqrt{2b^2cos^2x+c^2cos^2x-b^2(cos 2x)} dx

[/tex]

but i figured I should post here for ideas as well.