\((\triangle s)^{ 2 }=(x^{ 2 }+y^{ 2 }).(\triangle \theta )^{ 2 }\)
\((\triangle s)^{ 2 }=(x^{ 2 }+b^{ 2 }(1-\cfrac { x^{ 2 } }{ a^{ 2 } } )).(\triangle \theta )^{ 2 }\)
\((\triangle s)^{ 2 }=({ x }^{ 2 }(1-\cfrac { b^{ 2 } }{ a^{ 2 } } )+b^{ 2 }).(\triangle \theta )^{ 2 }\)
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 \(\triangle s\),
\((\triangle s)^{ 2 }=( a^2\cos ^{ 2 }{ \theta }(1-\cfrac { b^{ 2 } }{ a^{ 2 } } )+b^{ 2 }).(\triangle \theta )^{ 2 }\)
\(\iint {(ds)^2}=\iint { \cos ^{ 2 }{ \theta } ({ a }^{ 2 }-b^{ 2 })+{ b }^{ 2 } }. { (d\theta ) }^{ 2 }\)
\(\int {s(ds)}=\int {\frac{1}{2}( \theta +\sin( \theta )\cos( \theta ))({ a }^{ 2 }-b^{ 2 })+\theta{ b }^{ 2 }+A }. { d\theta }\)
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,
\( \cfrac { { s }^{ 2 } }{ 2 } =\cfrac { 1 }{ 4 } (\theta ^{ 2 }-{ \cos ^{ 2 }{ \theta } })({ a }^{ 2 }-b^{ 2 })+\cfrac { { \theta }^{ 2 } }{ 2 } { b }^{ 2 }-\cfrac { 1 }{ 2 } \pi ({ a }^{ 2 }+b^{ 2 })\theta |^{ 0 }_{ \pi }\)
\( =-\cfrac { 1 }{ 4 } (\pi ^{ 2 }-{ 1 })({ a }^{ 2 }-b^{ 2 })-\cfrac { { \pi }^{ 2 } }{ 2 } { b }^{ 2 }+\cfrac { 1 }{ 2 } { \pi }^{ 2 }({ a }^{ 2 }+b^{ 2 })-\cfrac { 1 }{ 4 } ({ a }^{ 2 }-b^{ 2 })\)
\( =-\cfrac { 1 }{ 4 } \pi ^{ 2 }({ a }^{ 2 }+b^{ 2 })+\cfrac { 1 }{ 2 } { \pi }^{ 2 }({ a }^{ 2 }+b^{ 2 })\)
\( =\cfrac { 1 }{ 4 } \pi ^{ 2 }({ a }^{ 2 }+b^{ 2 })\)
\( s\quad =\cfrac { \pi }{ \sqrt { 2 } } \sqrt { ({ a }^{ 2 }+b^{ 2 }) } \)
\( p=2s=2\pi \sqrt { \cfrac { ({ a }^{ 2 }+b^{ 2 }) }{ 2 } } \)
