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Question Number 60788 by ajfour last updated on 25/May/19

Commented by ajfour last updated on 25/May/19

$I n t e r m s o f α a n d e l l i p s e p a r a m e t e r s$ $a a n d b, f i n d m a x i m u m c o l o r e d a r e a .$
	Answered by mr W last updated on 27/May/19
	$α in diagram will be replaced by ∅. let μ=(a/b)≥1 (x^2 /a^2 )+(y^2 /b^2 )=1 ((r^2 cos^2 θ)/a^2 )+((r^2 sin^2 θ)/b^2 )=1 ⇒r^2 =((a^2 b^2 )/(a^2 sin^2 θ+b^2 cos^2 θ)) ((2x)/a^2 )+((2yy′)/b^2 )=0 ((r cos θ)/a^2 )+((r sin θy′)/b^2 )=0 ⇒y′=tan ϕ=−(a^2 /(b^2 tan θ))=−(μ^2 /(tan θ)) at A: r_A , θ_A =α, ϕ_A at B: r_B , θ_A =β, ϕ_B tan ϕ_A =−(a^2 /(b^2 tan α)) tan ϕ_B =−(a^2 /(b^2 tan β)) ∅=ϕ_A −ϕ_B tan ∅=tan (ϕ_A −ϕ_B )=((−(a^2 /(b^2 tan α))+(a^2 /(b^2 tan β)))/(1+(a^4 /(b^4 tan α tan β)))) ⇒tan ∅=((a/b))^2 ((tan α−tan β)/(((a/b))^4 +tan α tan β)) ⇒tan ∅=μ^2 ((tan α−tan β)/(μ^4 +tan α tan β)) A=shaded area A=∫_θ_A ^θ_B ((r^2 dθ)/2)−((r_A r_B sin (θ_B −θ_A ))/2) A=(1/2)∫_α ^β ((a^2 b^2 dθ)/(a^2 sin^2 θ+b^2 cos^2 θ))−((sin (β−α))/2)(√(((a^2 b^2 )/(a^2 sin^2 α+b^2 cos^2 α))×((a^2 b^2 )/(a^2 sin^2 β+b^2 cos^2 β)))) A=((a^2 b^2 )/2)∫_α ^β (dθ/(a^2 sin^2 θ+b^2 cos^2 θ))−((a^2 b^2 sin (β−α))/2)(√(1/((a^2 sin^2 α+b^2 cos^2 α)(a^2 sin^2 β+b^2 cos^2 β)))) A=((ab)/2)[tan^(−1) (((a tan β)/b))−tan^(−1) (((a tan α)/b))]−((a^2 b^2 sin (β−α))/(2b^2 ))(√(1/((μ^2 sin^2 α+cos^2 α)(μ^2 sin^2 β+cos^2 β)))) A=((ab)/2)[tan^(−1) (μ tan β)−tan^(−1) (μ tan α)]−((μ sin (β−α))/2)(√(1/((μ^2 sin^2 α+cos^2 α)(μ^2 sin^2 β+cos^2 β)))) A=((ab)/2){[tan^(−1) (μ tan β)−tan^(−1) (μ tan α)]−((μ sin (β−α))/(√((μ^2 sin^2 α+cos^2 α)(μ^2 sin^2 β+cos^2 β))))} =((ab)/2)×f ⇒f=tan^(−1) (μ tan β)−tan^(−1) (μ tan α)−((μ sin (β−α))/(√((μ^2 sin^2 α+cos^2 α)(μ^2 sin^2 β+cos^2 β)))) ⇒f=tan^(−1) [((μ(tan β−tan α) )/(1+μ^2 tan β tan α))]−((μ(tan β−tan α))/(√((1+μ^2 tan^2 α)(1+μ^2 tan^2 β)))) let p=μ tan α, q=μ tan β ⇒f=tan^(−1) (((q−p)/(1+qp)))−((q−p)/(√((p^2 +1)(q^2 +1)))) ...(i) tan ∅=μ^3 ×((μ tan α−μ tan β)/(μ^6 +μ^2 tan α tan β)) ((tan ∅)/μ^3 )=((p−q)/(μ^6 +pq )) tan ∅ μ^6 +tan ∅ pq=μ^3 p−μ^3 q ⇒q=((μ^3 (p−μ^3 tan ∅))/(μ^3 +p tan ∅)) ...(ii)$
	$α i n d i a g r a m w i l l b e r e p l a c e d b y \emptyset .$ $l e t μ = \frac{a}{b} ⩾ 1$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $\frac{r^{2} \cos^{2} θ}{a^{2}} + \frac{r^{2} \sin^{2} θ}{b^{2}} = 1$ $\Rightarrow r^{2} = \frac{a^{2} b^{2}}{a^{2} \sin^{2} θ + b^{2} \cos^{2} θ}$ $\frac{2 x}{a^{2}} + \frac{2 {y y}^{'}}{b^{2}} = 0$ $\frac{r \cos θ}{a^{2}} + \frac{r \sin θ y^{'}}{b^{2}} = 0$ $\Rightarrow y^{'} = \tan φ = - \frac{a^{2}}{b^{2} \tan θ} = - \frac{μ^{2}}{\tan θ}$ $a t A : r_{A}, θ_{A} = α, φ_{A}$ $a t B : r_{B}, θ_{A} = β, φ_{B}$ $\tan φ_{A} = - \frac{a^{2}}{b^{2} \tan α}$ $\tan φ_{B} = - \frac{a^{2}}{b^{2} \tan β}$ $\emptyset = φ_{A} - φ_{B}$ $\tan \emptyset = \tan (φ_{A} - φ_{B}) = \frac{- \frac{a^{2}}{b^{2} \tan α} + \frac{a^{2}}{b^{2} \tan β}}{1 + \frac{a^{4}}{b^{4} \tan α \tan β}}$ $\Rightarrow \tan \emptyset = {(\frac{a}{b})}^{2} \frac{\tan α - \tan β}{{(\frac{a}{b})}^{4} + \tan α \tan β}$ $\Rightarrow \tan \emptyset = μ^{2} \frac{\tan α - \tan β}{μ^{4} + \tan α \tan β}$ $A = s h a d e d a r e a$ $A = \int_{θ_{A}}^{θ_{B}} \frac{r^{2} d θ}{2} - \frac{r_{A} r_{B} \sin (θ_{B} - θ_{A})}{2}$ $A = \frac{1}{2} \int_{α}^{β} \frac{a^{2} b^{2} d θ}{a^{2} \sin^{2} θ + b^{2} \cos^{2} θ} - \frac{\sin (β - α)}{2} \sqrt{\frac{a^{2} b^{2}}{a^{2} \sin^{2} α + b^{2} \cos^{2} α} \times \frac{a^{2} b^{2}}{a^{2} \sin^{2} β + b^{2} \cos^{2} β}}$ $A = \frac{a^{2} b^{2}}{2} \int_{α}^{β} \frac{d θ}{a^{2} \sin^{2} θ + b^{2} \cos^{2} θ} - \frac{a^{2} b^{2} \sin (β - α)}{2} \sqrt{\frac{1}{(a^{2} \sin^{2} α + b^{2} \cos^{2} α) (a^{2} \sin^{2} β + b^{2} \cos^{2} β)}}$ $A = \frac{a b}{2} [\tan^{- 1} (\frac{a \tan β}{b}) - \tan^{- 1} (\frac{a \tan α}{b})] - \frac{a^{2} b^{2} \sin (β - α)}{2 b^{2}} \sqrt{\frac{1}{(μ^{2} \sin^{2} α + \cos^{2} α) (μ^{2} \sin^{2} β + \cos^{2} β)}}$ $A = \frac{a b}{2} [\tan^{- 1} (μ \tan β) - \tan^{- 1} (μ \tan α)] - \frac{μ \sin (β - α)}{2} \sqrt{\frac{1}{(μ^{2} \sin^{2} α + \cos^{2} α) (μ^{2} \sin^{2} β + \cos^{2} β)}}$ $A = \frac{a b}{2} {[\tan^{- 1} (μ \tan β) - \tan^{- 1} (μ \tan α)] - \frac{μ \sin (β - α)}{\sqrt{(μ^{2} \sin^{2} α + \cos^{2} α) (μ^{2} \sin^{2} β + \cos^{2} β)}}}$ $= \frac{a b}{2} \times f$ $\Rightarrow f = \tan^{- 1} (μ \tan β) - \tan^{- 1} (μ \tan α) - \frac{μ \sin (β - α)}{\sqrt{(μ^{2} \sin^{2} α + \cos^{2} α) (μ^{2} \sin^{2} β + \cos^{2} β)}}$ $\Rightarrow f = \tan^{- 1} [\frac{μ (\tan β - \tan α)}{1 + μ^{2} \tan β \tan α}] - \frac{μ (\tan β - \tan α)}{\sqrt{(1 + μ^{2} \tan^{2} α) (1 + μ^{2} \tan^{2} β)}}$ $l e t p = μ \tan α, q = μ \tan β$ $\Rightarrow f = \tan^{- 1} (\frac{q - p}{1 + q p}) - \frac{q - p}{\sqrt{(p^{2} + 1) (q^{2} + 1)}} . . . (i)$ $\tan \emptyset = μ^{3} \times \frac{μ \tan α - μ \tan β}{μ^{6} + μ^{2} \tan α \tan β}$ $\frac{\tan \emptyset}{μ^{3}} = \frac{p - q}{μ^{6} + p q}$ $\tan \emptyset μ^{6} + \tan \emptyset p q = μ^{3} p - μ^{3} q$ $\Rightarrow q = \frac{μ^{3} (p - μ^{3} \tan \emptyset)}{μ^{3} + p \tan \emptyset} . . . (i i)$
	Commented by ajfour last updated on 27/May/19

	$T h a n k s s o m u c h S i r, l e t m e$ $s o m e t i m e t o f o l l o w i t d e t e r m i n e d l y!$
	Commented by mr W last updated on 27/May/19

	Answered by ajfour last updated on 27/May/19
	$r=((ab)/(√(a^2 sin^2 θ+b^2 cos^2 θ))) A=∫_δ ^( δ+ξ) ((r^2 dθ)/2)−(1/2)r_A r_B sin ξ ((xcos δ)/a)+((ysin δ)/b)=1 ⇒ m_A =−(b/(atan δ)) m_B =−(b/(atan (δ+ξ))) tan α=((−(b/a)[(1/(tan δ))−(1/(tan (δ+ξ)))])/(1+(b^2 /(a^2 tan δtan (δ+ξ))))) ⇒ a^2 tan αtan δtan (δ+ξ)+b^2 tan α = ab[tan (δ+ξ)−tan δ] ⇒ tan (δ+ξ)=(b/a)(((btan α+atan δ)/(b−atan αtan δ))) ξ(δ)=tan^(−1) {(b/a)(((btan α+atan δ)/(b−atan αtan δ)))}−δ A=∫_δ ^( δ+ξ(δ)) ((r^2 dθ)/2)−(1/2)r_δ r_(δ+ξ(δ)) sin ξ(δ) (dA/dδ)=0 should yield a suitable 𝛅.$
	$r = \frac{a b}{\sqrt{a^{2} \sin^{2} θ + b^{2} \cos^{2} θ}}$ $A = \int_{δ}^{δ + ξ} \frac{r^{2} d θ}{2} - \frac{1}{2} r_{A} r_{B} \sin ξ$ $\frac{x \cos δ}{a} + \frac{y \sin δ}{b} = 1 \Rightarrow m_{A} = - \frac{b}{a \tan δ}$ $m_{B} = - \frac{b}{a \tan (δ + ξ)}$ $\tan α = \frac{- \frac{b}{a} [\frac{1}{\tan δ} - \frac{1}{\tan (δ + ξ)}]}{1 + \frac{b^{2}}{a^{2} \tan δ \tan (δ + ξ)}}$ $\Rightarrow a^{2} \tan α \tan δ \tan (δ + ξ) + b^{2} \tan α$ $= a b [\tan (δ + ξ) - \tan δ]$ $\Rightarrow \tan (δ + ξ) = \frac{b}{a} (\frac{b \tan α + a \tan δ}{b - a \tan α \tan δ})$ $ξ (δ) = \tan^{- 1} {\frac{b}{a} (\frac{b \tan α + a \tan δ}{b - a \tan α \tan δ})} - δ$ $A = \int_{δ}^{δ + ξ (δ)} \frac{r^{2} d θ}{2} - \frac{1}{2} r_{δ} r_{δ + ξ (δ)} \sin ξ (δ)$ $\frac{d A}{d δ} = 0 s h o u l d y i e l d a s u i t a b l e δ .$
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