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Question Number 142447 by ajfour last updated on 31/May/21

Commented by ajfour last updated on 31/May/21

$F i n d e l l i p s e p e r i m e t e r .$
	Answered by Dwaipayan Shikari last updated on 31/May/21

	$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $\Rightarrow y^{'} = - \frac{b^{2} x}{a^{2} y} = - \frac{b x}{a \sqrt{a^{2} - x^{2}}}$ $P e r i m e t r e = 2 \int_{- a}^{a} \sqrt{1 + {(y^{'})}^{2}} d x$ $= 4 \int_{0}^{a} \sqrt{1 + \frac{b^{2} x}{a^{2} (a^{2} - x^{2})}} d x x = u a$ $= 4 \int_{0}^{1} \frac{1}{a} \sqrt{\frac{a^{4} (1 - u^{2}) + a^{2} b^{2} u^{2}}{1 - u^{2}}} d u c o s θ = u$ $= 4 \int_{0}^{\frac{π}{2}} \sqrt{a^{2} {s i n}^{2} θ + b^{2} {c o s}^{2} θ} d θ$ $= 4 a \int_{0}^{\frac{π}{2}} \sqrt{1 - (1 - \frac{b^{2}}{a^{2}}) {c o s}^{2} θ} d θ$ $= 4 a \int_{0}^{\frac{π}{2}} \sqrt{1 - e^{2} {c o s}^{2} θ} d θ = 4 a \int_{0}^{\frac{π}{2}} \sqrt{1 - e^{2} {s i n}^{2} θ} d θ$ $= 4 a E (\frac{π}{2} ∣ e)$ $O r 4 a \int_{0}^{\frac{π}{2}} \sqrt{1 - e^{2} {s i n}^{2} θ} d θ = 4 a \underset{n = 0}{\sum^{\infty}} \frac{{(- \frac{1}{2})}_{n}}{n!} \int_{0}^{\frac{π}{2}} e^{2 n} {s i n}^{2 n} θ d θ$ $= 2 a \underset{n = 0}{\sum^{\infty}} \frac{{(- \frac{1}{2})}_{n} e^{2 n}}{n!} . \frac{Γ (n + \frac{1}{2}) Γ (\frac{1}{2})}{Γ (n + 1)}$ $= 2 π a \underset{n = 0}{\sum^{\infty}} \frac{{(- \frac{1}{2})}_{n} {(\frac{1}{2})}_{n}}{n! {(1)}_{n}} e^{2 n} = 2 π a_{2} F_{1} (- \frac{1}{2}, \frac{1}{2}; 1; e^{2})$
	Commented by Dwaipayan Shikari last updated on 31/May/21

	$A p p r o x i m a t i o n$ $4 a (\int_{0}^{\frac{π}{2}} 1 - \frac{1}{2} e^{2} {s i n}^{2} θ + O (e^{4})) \approx 4 a (\frac{π}{2} - \frac{e^{2} π}{8}) = 2 π a (1 - \frac{e^{2}}{4})$
	Answered by ajfour last updated on 01/Jun/21
	$let A be origin. (((x−a)^2 )/a^2 )+(y^2 /b^2 )=1 (((x−a))/a^2 )+((y((dy/dx)))/b^2 )=0 (dy/dx)=((b^2 (a−x))/(a^2 y)) dl=dx×(√(1+((b^2 (a−x)^2 )/(a^4 {1−(((x−a)^2 )/a^2 )})))) dl=dx(√(1+((b^2 (a−x)^2 )/(a^2 {a^2 −(a−x)^2 })))) =dx(√(1+((b^2 /a^2 )/({(1/((1−(x/a))^2 ))−1})))) let 1−(x/a)=(1/t) , (b/a)=m ⇒ −(dx/a)=−(dt/t^2 ) ⇒ dx=((adt)/t^2 ) dl=((adt)/t^2 )(√(1+(m^2 /(t^2 −1)))) let (m/( (√(t^2 −1))))=tan θ ⇒ t^2 −1=m^2 cot^2 θ tdt=−m^2 cot θcosec^2 θdθ dl=−((am^2 dθ)/(sin^3 θ(1+m^2 cot^2 θ)^(3/2) )) dl=−((am^2 dθ)/((sin^2 θ+m^2 cos^2 θ)^(3/2) )) let sin^2 θ+m^2 cos^2 θ=z^2 (1−m^2 )sin^2 θ=z^2 −m^2 or (1−m^2 )cos^2 θ=1−z^2 ⇒ (1−m^2 )sin θcos θdθ=zdz dθ=((zdz)/( (√(z^2 −m^2 ))(√(1−z^2 )))) dl=−((am^2 dz)/(z^2 (√(z^2 −m^2 ))(√(1−z^2 )))) let z=(√m)s ⇒ dz=(√m)ds dl=((−amds)/(s^2 (√(s^2 −m))(√(1−ms^2 )))) and if (1/s)=v dl=((amv^2 dv)/( (√(1−mv^2 ))(√(v^2 −m)))) dl=((amv^2 dv)/({(1+m^2 )v^2 −m(1+v^4 )}^(1/2) )) 2dl=((a(√m)d(v^2 ))/({(m+(1/m))−(v^2 +(1/v^2 ))}^(1/2) )) say v^2 =X , m+(1/m)=h^2 2dl=((a(√m)dX)/( (√(h^2 −(X+(1/X)))))) .... ........$
	$l e t A b e o r i g i n .$ $\frac{{(x - a)}^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $\frac{(x - a)}{a^{2}} + \frac{y (\frac{d y}{d x})}{b^{2}} = 0$ $\frac{d y}{d x} = \frac{b^{2} (a - x)}{a^{2} y}$ $d l = d x \times \sqrt{1 + \frac{b^{2} {(a - x)}^{2}}{a^{4} {1 - \frac{{(x - a)}^{2}}{a^{2}}}}}$ $d l = d x \sqrt{1 + \frac{b^{2} {(a - x)}^{2}}{a^{2} {a^{2} - {(a - x)}^{2}}}}$ $= d x \sqrt{1 + \frac{b^{2} / a^{2}}{{\frac{1}{{(1 - \frac{x}{a})}^{2}} - 1}}}$ $l e t 1 - \frac{x}{a} = \frac{1}{t}, \frac{b}{a} = m$ $\Rightarrow - \frac{d x}{a} = - \frac{d t}{t^{2}}$ $\Rightarrow d x = \frac{a d t}{t^{2}}$ $d l = \frac{a d t}{t^{2}} \sqrt{1 + \frac{m^{2}}{t^{2} - 1}}$ $l e t \frac{m}{\sqrt{t^{2} - 1}} = \tan θ$ $\Rightarrow t^{2} - 1 = m^{2} \cot^{2} θ$ $t d t = - m^{2} \cot θ cosec^{2} θ d θ$ $d l = - \frac{{a m}^{2} d θ}{\sin^{3} θ {(1 + m^{2} \cot^{2} θ)}^{3 / 2}}$ $d l = - \frac{{a m}^{2} d θ}{{(\sin^{2} θ + m^{2} \cos^{2} θ)}^{3 / 2}}$ $l e t \sin^{2} θ + m^{2} \cos^{2} θ = z^{2}$ $(1 - m^{2}) \sin^{2} θ = z^{2} - m^{2}$ $o r (1 - m^{2}) \cos^{2} θ = 1 - z^{2}$ $\Rightarrow (1 - m^{2}) \sin θ \cos θ d θ = z d z$ $d θ = \frac{z d z}{\sqrt{z^{2} - m^{2}} \sqrt{1 - z^{2}}}$ $d l = - \frac{{a m}^{2} d z}{z^{2} \sqrt{z^{2} - m^{2}} \sqrt{1 - z^{2}}}$ $l e t z = \sqrt{m} s \Rightarrow d z = \sqrt{m} d s$ $d l = \frac{- a m d s}{s^{2} \sqrt{s^{2} - m} \sqrt{1 - {m s}^{2}}}$ $a n d i f \frac{1}{s} = v$ $d l = \frac{{a m v}^{2} d v}{\sqrt{1 - {m v}^{2}} \sqrt{v^{2} - m}}$ $d l = \frac{{a m v}^{2} d v}{{(1 + m^{2}) v^{2} - m (1 + v^{4})}^{1 / 2}}$ $2 d l = \frac{a \sqrt{m} d (v^{2})}{{(m + \frac{1}{m}) - (v^{2} + \frac{1}{v^{2}})}^{1 / 2}}$ $s a y v^{2} = X, m + \frac{1}{m} = h^{2}$ $2 d l = \frac{a \sqrt{m} d X}{\sqrt{h^{2} - (X + \frac{1}{X})}}$ $. . . .$ $. . . . . . . .$
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