\(\triangle s\quad =\quad \sqrt { x^{ 2 }+y^{ 2 } } .\triangle \theta \)
\((\triangle s)^{ 2 }=(x^{ 2 }+y^{ 2 }).(\triangle \theta )^{ 2 }\)
\((\triangle s)^{ 2 }=(x^{ 2 }+b^{ 2 }(1-\cfrac { x^{ 2 } }{ a^{ 2 } } )).(\triangle \theta )^{ 2 }\)
\((\triangle s)^{ 2 }=({ x }^{ 2 }(1-\cfrac { b^{ 2 } }{ a^{ 2 } } )+b^{ 2 }).(\triangle \theta )^{ 2 }\)
Consider \(\cfrac { x }{ a } =cos(\theta )\) and \(\cfrac { y }{ b } =sin(\theta )\)
Since, \(sin^2(\theta )+ cos^2(\theta ) = 1= \cfrac { x^2 }{ a^2 }+\cfrac { y^2 }{ b^2 }\), which is equation of the ellipse.
Substituting \(x=acos(\theta )\) into the equation for \(\triangle s\),
\((\triangle s)^{ 2 }=( a^2\cos ^{ 2 }{ \theta }(1-\cfrac { b^{ 2 } }{ a^{ 2 } } )+b^{ 2 }).(\triangle \theta )^{ 2 }\)
\(\iint {(ds)^2}=\iint { \cos ^{ 2 }{ \theta } ({ a }^{ 2 }-b^{ 2 })+{ b }^{ 2 } }. { (d\theta ) }^{ 2 }\)
\(\int {s(ds)}=\int {\cfrac{1}{2}( \theta +\sin( \theta )\cos( \theta ))({ a }^{ 2 }-b^{ 2 })+\theta{ b }^{ 2 }+A }. { d\theta }\)
\( \cfrac{{ s }^{ 2 }}{2}=\cfrac { 1 }{ 4 } (\theta ^{ 2 }-{ \cos ^{ 2 }{ \theta } })({ a }^{ 2 }-b^{ 2 })+\cfrac { { \theta }^{ 2 } }{ 2 } { b }^{ 2 }+A\theta |^{ \cfrac { \pi }{ 2 } }_{ 0 }\)
\( \cfrac{{ s }^{ 2 } }{ 2 } =\cfrac { 1 }{ 4 } (\cfrac { \pi }{ 2 } )^{ 2 }({ a }^{ 2 }-b^{ 2 })+\cfrac { 1 }{ 2 } (\cfrac { \pi }{ 2 } )^{ 2 }{ b }^{ 2 }+A\cfrac { \pi }{ 2 } +\cfrac { 1 }{ 4 } ({ a }^{ 2 }-b^{ 2 })\)
\( { s }^{ 2 }=\cfrac { 1 }{ 2} (\cfrac { \pi }{ 2 } )^{ 2 }({ a }^{ 2 }-b^{ 2 })+(\cfrac { \pi }{ 2 } )^{ 2 }{ b }^{ 2 }+\cfrac { 1 }{ 2 } ({ a }^{ 2 }-b^{ 2 })+A\pi \)
\( { s }^{ 2 }=\cfrac { 1 }{ 2 } (\cfrac { \pi }{ 2 } )^{ 2 }{ a }^{ 2 }+\cfrac { 1 }{ 2 } (\cfrac { \pi }{ 2 } )^2{b}^{ 2 }+\cfrac { 1 }{ 2 } ({ a }^{ 2 }-{b}^{ 2 })+A\pi \)
\(s^2=\cfrac { 1 }{ 2 } [(\cfrac { \pi }{ 2 } )^{ 2 }{ a }^{ 2 }+(\cfrac { \pi }{ 2 } )^2{b}^{ 2 }+({ a }^{ 2 }-b^{ 2 })+2A\pi] \)
\(s=\cfrac { 1 }{ \sqrt{2} } \sqrt { (\cfrac { \pi }{ 2 } )^{ 2 }{ a }^{ 2 }+(\cfrac { \pi }{ 2 } )^2{b}^{ 2 }+({ a }^{ 2 }-b^{ 2 }) +2A\pi } \)
When \(a\) = \(b\), \(s\) = \(\cfrac{\pi}{2} a\), and swapping \(a\) and \(b\), \(s\) should remain the same.
These conditions should solve for \(A\).
When \(a\) = \(b\),
\(s=\cfrac { 1 }{ \sqrt{2} } \sqrt { (\cfrac { \pi }{ 2 } )^{ 2 }{ a }^{ 2 }+(\cfrac { \pi }{ 2 } )^2{a}^{ 2 }+2A\pi } \)
\(s=\cfrac { \pi }{ 2 } \cfrac { 1 }{ \sqrt{2} } \sqrt {2{ a }^{ 2 }+2A\pi } =\cfrac{\pi}{2}a\)
\(\therefore 2A\pi = 0\), therefore, \(A\) is of the form \(B(a^n-b^n)\)
on swapping \(a\) and \(b\),
\(({ a }^{ 2 }-b^{ 2 }) +2B(a^n-b^n)\pi = ({ b }^{ 2 }-a^{ 2 }) +2B(b^n-a^n)\pi \)
\(2({ a }^{ 2 }-b^{ 2 }+2B(a^n-b^n)\pi) = 0\)
\({ a }^{ 2 }-b^{ 2 }+2B(a^{ n }-b^{ n })\pi =0\)
\( { a }^{ 2 }+2B\pi a^{ n }-(b^{ 2 }+2B\pi b^{ n })=0\)
For the non-trivial case of \(a \neq b\), without loss of generality \(a>b\)
\( { a }^{ 2 }+2B\pi a^{ n }>(b^{ 2 }+2B\pi b^{ n })\) and so,
\( { a }^{ 2 }+2B\pi a^{ n }=0\) and \(b^{ 2 }+2B\pi b^{ n }=0\)
\(2B\pi b^{ n }=-{ b }^{ 2 }\)
\( B=-\cfrac { { b }^{ 2-n } }{ 2\pi } \) and
\(2B\pi a^{ n }=-{ a }^{ 2 }\)
\( B=-\cfrac { { a }^{ 2-n } }{ 2\pi }\)
This implies \(n\) = 2 and \(B=-\cfrac { 1 }{ 2\pi }\)
\(\therefore s=\cfrac { 1 }{ \sqrt{2} } \sqrt { (\cfrac { \pi }{ 2 } )^{ 2 }{ a }^{ 2 }+(\cfrac { \pi }{ 2 } )^2{b}^{ 2 }}\)
\(s=\cfrac { \pi }{ 2 \sqrt{2} } \sqrt { { a }^{ 2 }+{b}^{ 2 }}\)
For a full ellipse,
ellipse circumference = \( 4*s= 2\pi \sqrt { \cfrac{{ a }^{ 2 }+{b}^{ 2 }}{2}}\)
