Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 72408 by ajfour last updated on 28/Oct/19

Commented by ajfour last updated on 28/Oct/19

$A l l t h r e e c o l o u r e d r e g i o n s A, B, C$ $h a v e t h e s a m e a r e a . C e n t e r o f$ $c i r c l e l i e s o n m a j o r a x i s o f e l l i p s e .$ $L o c a t e t h e c e n t e r o f c i r c l e i n t e r m s$ $o f e l l i p s e p a r a m e t e r s a a n d b .$
	Answered by mr W last updated on 28/Oct/19

	Commented by mr W last updated on 29/Oct/19

	$e q n . o f c i r c l e :$ ${(x - h)}^{2} + y^{2} = r^{2}$ $e q n . o f e l l i p s e :$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $\frac{x^{2}}{a^{2}} + \frac{r^{2} - {(x - h)}^{2}}{b^{2}} = 1$ $b^{2} x^{2} + a^{2} r^{2} - a^{2} (x^{2} - 2 h x + h^{2}) = a^{2} b^{2}$ $(a^{2} - b^{2}) x^{2} - 2 a^{2} h x + a^{2} (b^{2} + h^{2} - r^{2}) = 0$ $x = \frac{a^{2} h - a \sqrt{a^{2} h^{2} - (a^{2} - b^{2}) (b^{2} + h^{2} - r^{2})}}{a^{2} - b^{2}}$ $x = \frac{a^{2} h - a \sqrt{(a^{2} - b^{2}) (r^{2} - b^{2}) + h^{2} b^{2}}}{a^{2} - b^{2}} = k$ $a r e a o f c i r c l e = a r e a o f e l l i p s e$ $π r^{2} = π a b$ $\Rightarrow r^{2} = a b$ $w i t h λ = \frac{b}{a}, η = \frac{h}{a}$ $\frac{k}{a} = κ = \frac{η - \sqrt{λ (1 - λ) (1 - λ^{2}) + λ^{2} η^{2}}}{1 - λ^{2}}$ $2 \int_{h - r}^{k} \sqrt{r^{2} - {(x - h)}^{2}} d x + 2 \int_{k}^{a} b \sqrt{1 - \frac{x^{2}}{a^{2}}} d x = \frac{π a b}{2}$ $\int_{h - \sqrt{λ} a}^{k} \sqrt{λ - {(\frac{x}{a} - \frac{h}{a})}^{2}} d (\frac{x}{a}) + λ \int_{k}^{a} \sqrt{1 - \frac{x^{2}}{a^{2}}} d (\frac{x}{a}) = \frac{π λ}{4}$ $\int_{η - \sqrt{λ}}^{κ} \sqrt{λ - {(ξ - η)}^{2}} d ξ + λ \int_{κ}^{1} \sqrt{1 - ξ^{2}} d ξ = \frac{π λ}{4}$ $\frac{1}{2} {[λ \sin^{- 1} \frac{ξ - η}{\sqrt{λ}} + (ξ - η) \sqrt{λ - {(ξ - η)}^{2}}]}_{η - \sqrt{λ}}^{κ} + \frac{λ}{2} {[\sin^{- 1} ξ + ξ \sqrt{1 - ξ^{2}}]}_{κ}^{1} = \frac{π λ}{4}$ $\Rightarrow \sin^{- 1} \frac{κ - η}{\sqrt{λ}} + \frac{κ - η}{\sqrt{λ}} \sqrt{1 - {(\frac{κ - η}{\sqrt{λ}})}^{2}} + \frac{π}{2} = \sin^{- 1} κ + κ \sqrt{1 - κ^{2}}$ $t h i s i s a n e q n . f o r η i n t e r m s o f λ .$ $e x a m p l e s :$ $λ = 0.5 \Rightarrow η = 0.6432$ $λ = 0.75 \Rightarrow η = 0.7441$ $λ = 2 \Rightarrow η = 0.9096$
	Commented by ajfour last updated on 28/Oct/19

	$T h a n k s s i r . . .$ $I t h i n k, a l l t h a t s p o s s i b l e, y o u$ $d i d i t!$
	Commented by mr W last updated on 28/Oct/19

	Commented by mr W last updated on 29/Oct/19

	Commented by ajfour last updated on 29/Oct/19

	$E x c e l l e n t s i r! h o w a b o u t t h e 3 D$ $a n a l o g u e w h e n t h e t h r e e$ $c o m p a r t m e n t v o l u m e s b e e q u a l ?$ $i j u s t w o n d e r . .$
	Commented by ajfour last updated on 29/Oct/19
	$x=h−rcos θ , y=rsin θ , r=(√(ab)) (((h−rcos θ)^2 )/a^2 )+((r^2 sin^2 θ)/b^2 )=1 let cos θ=t ⇒ b^2 (h^2 −2hrt+r^2 t^2 )+a^2 r^2 (1−t^2 ) =a^2 b^2 (a^2 −b^2 )r^2 t^2 +2b^2 hrt−(b^2 h^2 +a^2 r^2 )=0 t=((−b^2 hr+(√(b^4 h^2 r^2 +(a^2 −b^2 )(b^2 h^2 +a^2 r^2 ))))/((a^2 −b^2 )r^2 )) ellipse y=(b/a)(√(a^2 −x^2 )) A=(b/a)∫_(−a) ^( h−rt) (√(a^2 −x^2 ))dx−((r^2 θ)/2)+((r^2 sin θcos θ)/2) = ((πab)/4) {(x/2)(√(a^2 −x^2 ))+(a^2 /2)sin^(−1) (x/a)}_(−a) ^(h−rt) −((r^2 θ)/2)+((r^2 sin θcos θ)/2)=((πab)/4) ⇒ (((h−rt)/2))(√(a^2 −(h−rt)^2 ))+(a^2 /2)sin^(−1) (((h−rt)/a))+((πa^2 )/4) −((r^2 cos^(−1) t)/2)+((r^2 t(√(1−t^2 )))/2)=((πab)/4) t=cos θ is obtained from above eq. Then to find h, t=((−b^2 hr+(√(b^4 h^2 r^2 +(a^2 −b^2 )(b^2 h^2 +a^2 r^2 ))))/((a^2 −b^2 )r^2 )) ..$
	$x = h - r \cos θ, y = r \sin θ, r = \sqrt{a b}$ $\frac{{(h - r \cos θ)}^{2}}{a^{2}} + \frac{r^{2} \sin^{2} θ}{b^{2}} = 1$ $l e t \cos θ = t$ $\Rightarrow b^{2} (h^{2} - 2 h r t + r^{2} t^{2}) + a^{2} r^{2} (1 - t^{2})$ $= a^{2} b^{2}$ $(a^{2} - b^{2}) r^{2} t^{2} + 2 b^{2} h r t - (b^{2} h^{2} + a^{2} r^{2}) = 0$ $t = \frac{- b^{2} h r + \sqrt{b^{4} h^{2} r^{2} + (a^{2} - b^{2}) (b^{2} h^{2} + a^{2} r^{2})}}{(a^{2} - b^{2}) r^{2}}$ $e l l i p s e$ $y = \frac{b}{a} \sqrt{a^{2} - x^{2}}$ $A = \frac{b}{a} \int_{- a}^{h - r t} \sqrt{a^{2} - x^{2}} d x - \frac{r^{2} θ}{2} + \frac{r^{2} \sin θ \cos θ}{2}$ $= \frac{π a b}{4}$ ${\frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} \frac{x}{a}}_{- a}^{h - r t}$ $- \frac{r^{2} θ}{2} + \frac{r^{2} \sin θ \cos θ}{2} = \frac{π a b}{4}$ $\Rightarrow (\frac{h - r t}{2}) \sqrt{a^{2} - {(h - r t)}^{2}} + \frac{a^{2}}{2} \sin^{- 1} (\frac{h - r t}{a}) + \frac{π a^{2}}{4}$ $- \frac{r^{2} \cos^{- 1} t}{2} + \frac{r^{2} t \sqrt{1 - t^{2}}}{2} = \frac{π a b}{4}$ $t = \cos θ i s o b t a i n e d f r o m a b o v e e q .$ $T h e n t o f i n d h,$ $t = \frac{- b^{2} h r + \sqrt{b^{4} h^{2} r^{2} + (a^{2} - b^{2}) (b^{2} h^{2} + a^{2} r^{2})}}{(a^{2} - b^{2}) r^{2}}$ $. .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum