Ellipse 2014

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

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

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

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

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 $\triangle s$,

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

$\iint {(ds)^2}=\iint { \cos ^{ 2 }{ \theta  } ({ a }^{ 2 }-b^{ 2 })+{ b }^{ 2 } }. { (d\theta ) }^{ 2 }$

$\int {s(ds)}=\int {\frac{1}{2}( \theta +\sin( \theta )\cos( \theta ))({ a }^{ 2 }-b^{ 2 })+\theta{ b }^{ 2 }+A }. { d\theta }$

$s\cfrac { ds }{ d\theta  } =\cfrac { 1 }{ 2 } (\theta +\sin  (\theta )\cos  (\theta ))({ a }^{ 2 }-b^{ 2 })+\theta { b }^{ 2 }+A$

When $\theta =\pi$,$s=0$,   so $s\cfrac { ds }{ d\theta  }=0$  since $A$ is independent of $\theta$ and not infinity (infinity is not a constant).

$\therefore  \cfrac { 1 }{ 2 } \pi ({ a }^{ 2 }-b^{ 2 })+\pi { b }^{ 2 }+A=0$

$A=-\cfrac { 1 }{ 2 } \pi ({ a }^{ 2 }+b^{ 2 })$,    so

$\int {s(ds)}=\int {\cfrac { 1 }{ 2 } (\theta +\sin  (\theta )\cos  (\theta ))({ a }^{ 2 }-b^{ 2 })+\theta { b }^{ 2 }-\cfrac { 1 }{ 2 } \pi ({ a }^{ 2 }+b^{ 2 })} { d\theta }$

Formulating the definite integral in the positive $x$ direction,

$\cfrac { { s }^{ 2 } }{ 2 } =\cfrac { 1 }{ 4 } (\theta ^{ 2 }-{ \cos ^{ 2 }{ \theta  }  })({ a }^{ 2 }-b^{ 2 })+\cfrac { { \theta  }^{ 2 } }{ 2 } { b }^{ 2 }-\cfrac { 1 }{ 2 } \pi ({ a }^{ 2 }+b^{ 2 })\theta |^{ 0 }_{ \pi  }$

$=-\cfrac { 1 }{ 4 } (\pi ^{ 2 }-{ 1 })({ a }^{ 2 }-b^{ 2 })-\cfrac { { \pi  }^{ 2 } }{ 2 } { b }^{ 2 }+\cfrac { 1 }{ 2 } { \pi  }^{ 2 }({ a }^{ 2 }+b^{ 2 })-\cfrac { 1 }{ 4 } ({ a }^{ 2 }-b^{ 2 })$

$=-\cfrac { 1 }{ 4 } \pi ^{ 2 }({ a }^{ 2 }+b^{ 2 })+\cfrac { 1 }{ 2 } { \pi  }^{ 2 }({ a }^{ 2 }+b^{ 2 })$

$=\cfrac { 1 }{ 4 } \pi ^{ 2 }({ a }^{ 2 }+b^{ 2 })$

$s\quad =\cfrac { \pi  }{ \sqrt { 2 }  } \sqrt { ({ a }^{ 2 }+b^{ 2 }) }$

$p=2s=2\pi \sqrt { \cfrac { ({ a }^{ 2 }+b^{ 2 }) }{ 2 }  }$