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Question Number 102980 by ajfour last updated on 12/Jul/20

Commented by ajfour last updated on 12/Jul/20

$F i n d a / b i f r e g i o n s 1, 2, 3, 4, 5$ $h a v e e q u a l a r e a s .$
Commented by mr W last updated on 12/Jul/20

$t h i s i s n o t p o s s i b l e f o r e l l i p s e$ $n o m a t t e r w h i c h v a l u e a / b h a s .$
Commented by mr W last updated on 12/Jul/20

$i f t h i s i s t r u e, s a y f o r a c e r t a i n$ $v a l u e o f a / b, t h e n i t m u s t b e a l s o t r u e$ $w h e n w e s h r i n k t h e e l l i p s e i n$ $x - d i r e c t i o n i n r a t i o b / a t o a c i r c l e$ $w i t h r a d i u s r = b .$ $t h a t m e a n s e a c h o f t h e t h r e e s h a d e d$ $s e g m e n t s o f t h e c i r c l e m u s t h a v e$ $f i f t h o f t h e a r e a o f t h e c i r c l e . b u t$ $t h i s i s n o t p o s s i b l e, s i n c e$ $a r e a o f s e g e m e n t = \frac{b^{2}}{2} (\frac{2 π}{3} - \frac{\sqrt{3}}{2}) \neq \frac{π b^{2}}{5}$
Commented by mr W last updated on 12/Jul/20

Commented by ajfour last updated on 12/Jul/20

$^{″} t h e n i t m u s t b e a l s o t r u e$ $w h e n w e s h r i n k t h e e l l i p s e^{″}$ $m a y n o t n e c e s s a r i l y b e t r u e$ $S i r . . . .$
Commented by mr W last updated on 12/Jul/20

$w h e n a n e l l i p s e i s s h r i n k e d i n$ $x - d i r e c t i o n i n r a t i o \frac{b}{a}, t h e a r e a o f$ $e a c h p a r t i s s h r i n k e d i n t h e s a m e$ $r a t i o \frac{b}{a} :$ $A_{1} \to A_{1}^{^{'}} = \frac{b}{a} A_{1}$ $A_{2} \to A_{2}^{^{'}} = \frac{b}{a} A_{2}$ $. . .$ $t h e r e f o r e i f A_{1} = A_{2} = A_{3} = . . ., w e g e t$ $a l s o A_{1}^{'} = A_{2}^{'} = A_{3}^{'} = . . .$
	Answered by ajfour last updated on 12/Jul/20
	$A_1 =((2b)/a)∫_h ^( a) (√(a^2 −x^2 )) dx = ((2b)/a){(h/2)(√(a^2 −h^2 ))+(a^2 /2)sin^(−1) (h/a)}∣_h ^a =((2b)/a){((πa^2 )/4)−(h/2)(√(a^2 −h^2 ))−(a^2 /2)sin^(−1) (h/a)} A_2 =(k/2)(h+a) , k=b(√(1−(h^2 /a^2 ))) A_1 = A_2 =((πab)/5) gives h and (a/b) ((2b)/a){((πa^2 )/4)−(h/2)(√(a^2 −h^2 ))−(a^2 /2)sin^(−1) (h/a)} = ((b(h+a))/2)(√(1−(h^2 /a^2 ))) let (h/a)= p ⇒ (π/2)−p(√(1−p^2 ))−sin^(−1) p = (((p+1))/2)(√(1−p^2 )) = (π/5) Yes Sir, follow you only now!$
	$A_{1} = \frac{2 b}{a} \int_{h}^{a} \sqrt{a^{2} - x^{2}} d x$ $= \frac{2 b}{a} {\frac{h}{2} \sqrt{a^{2} - h^{2}} + \frac{a^{2}}{2} \sin^{- 1} \frac{h}{a}} ∣_{h}^{a}$ $= \frac{2 b}{a} {\frac{π a^{2}}{4} - \frac{h}{2} \sqrt{a^{2} - h^{2}} - \frac{a^{2}}{2} \sin^{- 1} \frac{h}{a}}$ $A_{2} = \frac{k}{2} (h + a), k = b \sqrt{1 - \frac{h^{2}}{a^{2}}}$ $A_{1} = A_{2} = \frac{π a b}{5} g i v e s h a n d \frac{a}{b}$ $\frac{2 b}{a} {\frac{π a^{2}}{4} - \frac{h}{2} \sqrt{a^{2} - h^{2}} - \frac{a^{2}}{2} \sin^{- 1} \frac{h}{a}}$ $= \frac{b (h + a)}{2} \sqrt{1 - \frac{h^{2}}{a^{2}}}$ $l e t \frac{h}{a} = p$ $\Rightarrow \frac{π}{2} - p \sqrt{1 - p^{2}} - \sin^{- 1} p$ $= \frac{(p + 1)}{2} \sqrt{1 - p^{2}} = \frac{π}{5}$ $Y e s S i r, f o l l o w y o u o n l y n o w!$
	Commented by mr W last updated on 12/Jul/20

	$y o u c a n m a k e A_{1} = A_{2} o r A_{1} = A_{3} o r$ $A_{2} = A_{3} f o r a n y e l l i p s e, b u t n e v e r$ $A_{1} = A_{2} = A_{3} .$
	Commented by ajfour last updated on 12/Jul/20

	$A_{1} = A_{2} g i v e s h f o r a n y a a n d b;$ $w h i l e A_{1} = A_{2} = \frac{π a b}{5} f o r s p e c i f i c$ $\frac{a}{b} .$
	Commented by mr W last updated on 12/Jul/20

	$y e s, y o u c a n g e t A_{1} = A_{2} f o r a n y$ $e l l i p s e, b u t n e v e r A_{1} = A_{2} = \frac{π a b}{5},$ $b e c a u s e A_{1} = A_{2} = \frac{π a b}{5} m e a n s t h e s a m e$ $a s A_{1} = A_{2} = A_{3} = \frac{π a b}{5}, w h i c h i s n o t$ $p o s s i b l e .$
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