Ellipse 1986

$\triangle s\quad =\quad \sqrt { x^{ 2 }+y^{ 2 } } .\triangle \theta$

$(\triangle s)^{ 2 }=(x^{ 2 }+y^{ 2 }).(\triangle \theta )^{ 2 }$

$(\triangle s)^{ 2 }=(x^{ 2 }+b^{ 2 }(1-\cfrac { x^{ 2 } }{ a^{ 2 } } )).(\triangle \theta )^{ 2 }$

$(\triangle s)^{ 2 }=({ x }^{ 2 }(1-\cfrac { b^{ 2 } }{ a^{ 2 } } )+b^{ 2 }).(\triangle \theta )^{ 2 }$

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 = $4*s= 2\pi  \sqrt { \cfrac{{ a }^{ 2 }+{b}^{ 2 }}{2}}$