Find the circumference of an ellipse (((x−2)^2 )/(16))+(y^2 /(25)) = 1


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Question Number 132402 by bramlexs22 last updated on 14/Feb/21

$Find the circumference of an$ $ellipse \frac{{(x - 2)}^{2}}{16} + \frac{y^{2}}{25} = 1$
Commented by mr W last updated on 14/Feb/21

$t h e p e r i m e t e r o f a n e l l i p s e c a n n o t$ $b e e x a c t l y c a l c u l a t e d .$
Commented by mr W last updated on 14/Feb/21

Commented by bramlexs22 last updated on 14/Feb/21

$thank you$
Commented by Dwaipayan Shikari last updated on 14/Feb/21

$S i r, d o y o u h a v e a n y i d e a a b o u t t h e d e r i v a t i o n {R a m a n u j a n}^{'} s$ $a p p r o x i m a t i o n ?$
Commented by mr W last updated on 14/Feb/21

$n o . t h e g e n i u s {d i d n}^{'} t t e l l h o w h e$ $c a m e t o t h e f o r m u l a .$
	Answered by EDWIN88 last updated on 14/Feb/21
	$the equation of ellipse (((x−2)^2 )/4^2 ) + (y^2 /5^2 ) = 1 by using formula C = π[ 3(a+b)−(√((3a+b)(a+3b))) ] in this case { ((a=4)),((b=5)) :} ⇒ C = π [ 3(9)−(√((12+5)(4+15))) ] C = π [ 27−(√(17×19)) ] ≈ 28.362$
	$the equation of ellipse \frac{{(x - 2)}^{2}}{4^{2}} + \frac{y^{2}}{5^{2}} = 1$ $by using formula C = π [3 (a + b) - \sqrt{(3 a + b) (a + 3 b)}]$ $in this case {\begin{cases} a = 4 \\ b = 5 \end{cases} \Rightarrow C = π [3 (9) - \sqrt{(12 + 5) (4 + 15)}]$ $C = π [27 - \sqrt{17 \times 19}] \approx 28.362$
	Answered by MJS_new last updated on 14/Feb/21

	$length of curve y = c (x) with α ⩽ x ⩽ β we have$ $L = \underset{α}{\int^{β}} \sqrt{1 + {(c^{'} (x))}^{2}} d x$ $for y = \pm \frac{b}{a} \sqrt{a^{2} - x^{2}}$ $L = \frac{2}{a} \underset{- a}{\int^{a}} \sqrt{\frac{a^{4} - (a^{2} - b^{2}) x^{2}}{a^{2} - x^{2}}} d x =$ $[t = \arcsin \frac{x}{a} \to d x = \sqrt{a^{2} - x^{2}} d t]$ $= 2 a \underset{- π / 2}{\int^{π / 2}} \sqrt{1 - \frac{a^{2} - b^{2}}{a^{2}} \sin^{2} t} d t =$ $= 2 a {[E (x ∣ \frac{a^{2} - b^{2}}{a^{2}})]}_{- π / 2}^{π / 2}$
	Commented by MJS_new last updated on 14/Feb/21
	I get approximately 28.3616678890
	Answered by Dwaipayan Shikari last updated on 14/Feb/21

	$P e r i m e t e r$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \Rightarrow \frac{x}{a^{2}} = - \frac{y}{b^{2}} y^{'} \Rightarrow y^{'} = - \frac{b x}{a \sqrt{a^{2} - x^{2}}}$ $A r c l e n g t h = \int_{- a}^{a} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x$ $P e r i m e t e r = 2 \int_{- a}^{a} \sqrt{1 + \frac{b^{2} x^{2}}{a^{2} (a^{2} - x^{2})}} d x$ $= \frac{2}{a} \int_{- a}^{a} \sqrt{\frac{a^{4} - a^{2} x^{2} + b^{2} x^{2}}{a^{2} - x^{2}}} x = a u$ $= 2 \int_{- 1}^{1} \sqrt{\frac{a^{2} (a^{2} - a^{2} u^{2} + b^{2} u^{2})}{a^{2} (1 - u^{2})}} d u$ $= 2 \int_{- \frac{π}{2}}^{\frac{π}{2}} \sqrt{a^{2} {s i n}^{2} θ + b^{2} {c o s}^{2} θ} d θ$ $= 2 a \int_{- \frac{π}{2}}^{\frac{π}{2}} \sqrt{\frac{b^{2}}{a^{2}} + e^{2} {s i n}^{2} θ} d θ = 2 b \int_{- \frac{π}{2}}^{\frac{π}{2}} \sqrt{1 + \frac{a^{2} e^{2}}{b^{2}} {s i n}^{2} θ} d θ$ $= 4 b E (\frac{π}{2} ∣ \frac{a^{2} e^{2}}{b^{2}}) (E l l i p t i c i n t e g r a l o f f i r s t k i n d)$
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