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Question Number 177628 by mr W last updated on 07/Oct/22

Commented by mr W last updated on 07/Oct/22

$e l i p s e w i t h a = 10 c m a n d b = 6 c m i s$ $r o t a t e d a b o u t i t s c e n t e r b y 60 ° . f i n d$ $t h e s h a d e d a r e a .$
	Answered by mr W last updated on 08/Oct/22

	Commented by mr W last updated on 09/Oct/22
	$red elipse: (x^2 /(10^2 ))+(y^2 /6^2 )=1 ((r^2 cos^2 θ)/(10^2 ))+((y^2 sin^2 θ)/6^2 )=1 ⇒r=((60)/( (√(36 cos^2 θ+100 sin^2 θ)))) blue elipse: ⇒r=((60)/( (√(36 cos^2 (θ−ϕ)+100 sin^2 (θ−ϕ))))) intersection point P, Q: ((60)/( (√(36 cos^2 (θ−ϕ)+100 sin^2 (θ−ϕ)))))=((60)/( (√(36 cos^2 θ+100 sin^2 θ)))) 36 cos^2 (θ−ϕ)+100 sin^2 (θ−ϕ)=36 cos^2 θ+100 sin^2 θ 9 [cos^2 (θ−ϕ)−cos^2 θ]+25 [sin^2 (θ−ϕ)−sin^2 θ]=0 9 [cos (θ−ϕ)+cos θ][cos (θ−ϕ)−cos θ]+25 [sin (θ−ϕ)+sin θ][sin (θ−ϕ)−sin θ]=0 9(cos θ cos ϕ+sin θ sin ϕ+cos θ)(cos θ cos ϕ+sin θ sin ϕ−cos θ)+25 (sin θ cos ϕ−cos θ sin ϕ+sin θ)(sin θ cos ϕ−cos θ sin ϕ−sin θ)=0 9(((1+cos ϕ)/(sin ϕ))+tan θ)(((1−cos ϕ)/(sin ϕ))−tan θ)+25(((1+cos ϕ)/(sin ϕ))tan θ−1)(((1−cos ϕ)/(sin ϕ))tan θ+1)=0 tan^2 θ+((2 tan θ)/(tan ϕ))−1=0 with ϕ=60° tan^2 θ+((2 tan θ)/( (√3)))−1=0 tan θ=((−1±2)/( (√3)))= { ((1/( (√3)))),((−(√3))) :} ⇒θ_Q =(π/6) ⇒θ_P =π−(π/3)=((2π)/3) area of sector of blue elipse: A_(blue sector) =(1/2)∫_(θ_Q −ϕ) ^(θ_P −ϕ) r^2 dθ area of sector of red elipse: A_(red sector) =(1/2)∫_θ_Q ^θ_P r^2 dθ shaded are=2(sector blue−sector red): A_(shaded) =∫_(θ_Q −ϕ) ^(θ_P −ϕ) r^2 dθ−∫_θ_Q ^θ_P r^2 dθ A_(shaded) =∫_(θ_Q −ϕ) ^θ_Q r^2 dθ−∫_(θ_P −ϕ) ^θ_P r^2 dθ ∫(dθ/(b^2 cos^2 θ+a^2 sin^2 θ))=(1/(ab))×tan^(−1) ((a/b)×tan θ)+C ∫r^2 dθ=∫((3600 dθ)/( 36 cos^2 θ+100 sin^2 θ)) =60 tan^(−1) (((5 tan θ)/3))+C (A_(shadex) /(60))=[tan^(−1) (((5 tan θ)/3))]_(θ_Q −(π/3)) ^θ_Q −[tan^(−1) (((5 tan θ)/3))]_(θ_P −(π/3)) ^θ_P (A_(shadex) /(60))=[tan^(−1) (((5 tan θ)/3))]_(−(π/6)) ^(π/6) −[tan^(−1) (((5 tan θ)/3))]_(π/3) ^((2π)/3) (A_(shadex) /(60))=2 tan^(−1) ((5/( 3(√3))))−[π−2 tan^(−1) ((5/( (√3))))] (A_(shadex) /(60))=2[tan^(−1) ((5/(3(√3))))+tan^(−1) ((5/( (√3))))]−π A_(shaded) =120[tan^(−1) ((5/(3(√3))))+tan^(−1) ((5/( (√3))))]−60π ≈51.9227$
	$r e d e l i p s e :$ $\frac{x^{2}}{10^{2}} + \frac{y^{2}}{6^{2}} = 1$ $\frac{r^{2} \cos^{2} θ}{10^{2}} + \frac{y^{2} \sin^{2} θ}{6^{2}} = 1$ $\Rightarrow r = \frac{60}{\sqrt{36 \cos^{2} θ + 100 \sin^{2} θ}}$ $b l u e e l i p s e :$ $\Rightarrow r = \frac{60}{\sqrt{36 \cos^{2} (θ - φ) + 100 \sin^{2} (θ - φ)}}$ $i n t e r s e c t i o n p o i n t P, Q :$ $\frac{60}{\sqrt{36 \cos^{2} (θ - φ) + 100 \sin^{2} (θ - φ)}} = \frac{60}{\sqrt{36 \cos^{2} θ + 100 \sin^{2} θ}}$ $36 \cos^{2} (θ - φ) + 100 \sin^{2} (θ - φ) = 36 \cos^{2} θ + 100 \sin^{2} θ$ $9 [\cos^{2} (θ - φ) - \cos^{2} θ] + 25 [\sin^{2} (θ - φ) - \sin^{2} θ] = 0$ $9 [\cos (θ - φ) + \cos θ] [\cos (θ - φ) - \cos θ] + 25 [\sin (θ - φ) + \sin θ] [\sin (θ - φ) - \sin θ] = 0$ $9 (\cos θ \cos φ + \sin θ \sin φ + \cos θ) (\cos θ \cos φ + \sin θ \sin φ - \cos θ) + 25 (\sin θ \cos φ - \cos θ \sin φ + \sin θ) (\sin θ \cos φ - \cos θ \sin φ - \sin θ) = 0$ $9 (\frac{1 + \cos φ}{\sin φ} + \tan θ) (\frac{1 - \cos φ}{\sin φ} - \tan θ) + 25 (\frac{1 + \cos φ}{\sin φ} \tan θ - 1) (\frac{1 - \cos φ}{\sin φ} \tan θ + 1) = 0$ $\tan^{2} θ + \frac{2 \tan θ}{\tan φ} - 1 = 0$ $w i t h φ = 60 °$ $\tan^{2} θ + \frac{2 \tan θ}{\sqrt{3}} - 1 = 0$ $\tan θ = \frac{- 1 \pm 2}{\sqrt{3}} = {\begin{cases} \frac{1}{\sqrt{3}} \\ - \sqrt{3} \end{cases}$ $\Rightarrow θ_{Q} = \frac{π}{6}$ $\Rightarrow θ_{P} = π - \frac{π}{3} = \frac{2 π}{3}$ $a r e a o f s e c t o r o f b l u e e l i p s e :$ $A_{b l u e s e c t o r} = \frac{1}{2} \int_{θ_{Q} - φ}^{θ_{P} - φ} r^{2} d θ$ $a r e a o f s e c t o r o f r e d e l i p s e :$ $A_{r e d s e c t o r} = \frac{1}{2} \int_{θ_{Q}}^{θ_{P}} r^{2} d θ$ $s h a d e d a r e = 2 (s e c t o r b l u e - s e c t o r r e d) :$ $A_{s h a d e d} = \int_{θ_{Q} - φ}^{θ_{P} - φ} r^{2} d θ - \int_{θ_{Q}}^{θ_{P}} r^{2} d θ$ $A_{s h a d e d} = \int_{θ_{Q} - φ}^{θ_{Q}} r^{2} d θ - \int_{θ_{P} - φ}^{θ_{P}} r^{2} d θ$ $\int \frac{d θ}{b^{2} \cos^{2} θ + a^{2} \sin^{2} θ} = \frac{1}{a b} \times \tan^{- 1} (\frac{a}{b} \times \tan θ) + C$ $\int r^{2} d θ = \int \frac{3600 d θ}{36 \cos^{2} θ + 100 \sin^{2} θ}$ $= 60 \tan^{- 1} (\frac{5 \tan θ}{3}) + C$ $\frac{A_{s h a d e x}}{60} = {[\tan^{- 1} (\frac{5 \tan θ}{3})]}_{θ_{Q} - \frac{π}{3}}^{θ_{Q}} - {[\tan^{- 1} (\frac{5 \tan θ}{3})]}_{θ_{P} - \frac{π}{3}}^{θ_{P}}$ $\frac{A_{s h a d e x}}{60} = {[\tan^{- 1} (\frac{5 \tan θ}{3})]}_{- \frac{π}{6}}^{\frac{π}{6}} - {[\tan^{- 1} (\frac{5 \tan θ}{3})]}_{\frac{π}{3}}^{\frac{2 π}{3}}$ $\frac{A_{s h a d e x}}{60} = 2 \tan^{- 1} (\frac{5}{3 \sqrt{3}}) - [π - 2 \tan^{- 1} (\frac{5}{\sqrt{3}})]$ $\frac{A_{s h a d e x}}{60} = 2 [\tan^{- 1} (\frac{5}{3 \sqrt{3}}) + \tan^{- 1} (\frac{5}{\sqrt{3}})] - π$ $A_{s h a d e d} = 120 [\tan^{- 1} (\frac{5}{3 \sqrt{3}}) + \tan^{- 1} (\frac{5}{\sqrt{3}})] - 60 π$ $\approx 51.9227$
	Commented by a.lgnaoui last updated on 08/Oct/22

	$B o n n e i n t e r p r e t a t i o n M e r c i .$
	Commented by mr W last updated on 09/Oct/22

	Commented by mr W last updated on 09/Oct/22

	$a l t e r n a t i v e :$ $A_{s h a d e d} = A_{e l i p s e} - 4 A_{r e d s e c t o r}$ $= 60 π - 2 \times 60 {[\tan^{- 1} (\frac{5 \tan θ}{3})]}_{\frac{π}{6}}^{\frac{2 π}{3}}$ $= 60 π - 2 \times 60 [π - \tan^{- 1} (\frac{5 \sqrt{3}}{3}) - \tan^{- 1} (\frac{5}{3 \sqrt{3}})]$ $= 120 [\tan^{- 1} (\frac{5 \sqrt{3}}{3}) + \tan^{- 1} (\frac{5}{3 \sqrt{3}})] - 60 π$ $\approx 51.9227$
	Commented by Tawa11 last updated on 09/Oct/22

	$Great sir$
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