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Question Number 63108 by ajfour last updated on 29/Jun/19

Commented by ajfour last updated on 29/Jun/19

$F i n d x_{A} i n t e r m s o f a, b, s .$ $s_{m i n} ⩽ s ⩽ s_{m a x} .$ $a, b a r e p a r a m e t e r s o f e l l i p s e$ $w h i l e s i s s i d e l e n g t h o f △_{e q u i l a t e r a l} .$
	Answered by mr W last updated on 02/Jul/19

	$R = \frac{s}{\sqrt{3}} = r a d i u s o f c i r c u m c i r c l e$ $M (h, k) = c e n t r o i d a n d c i r c u m c e n t e r$ $l e t p = \frac{h}{a}, q = \frac{k}{b}, α = \frac{R}{a} = \frac{s}{\sqrt{3} a}, β = \frac{R}{b} = \frac{s}{\sqrt{3} b}$ $x_{A} = h + R \cos θ = a (\frac{h}{a} + \frac{R}{a} \cos θ) = a (p + α \cos θ)$ $y_{A} = k + R \sin θ = b (\frac{k}{b} + \frac{R}{b} \sin θ) = b (q + β \sin θ)$ $\frac{x_{A}^{2}}{a^{2}} + \frac{y_{A}^{2}}{b^{2}} = 1$ $\Rightarrow {(p + α \cos θ)}^{2} + {(q + β \sin θ)}^{2} = 1$ $\Rightarrow p^{2} + 2 α p \cos θ + α^{2} \cos^{2} θ + q^{2} + 2 β q \sin θ + β^{2} \sin^{2} θ = 1 . . . (i)$ $x_{B} = a [p + α \cos (θ + \frac{2 π}{3})]$ $y_{B} = b [q + β \sin (θ + \frac{2 π}{3})]$ $\Rightarrow {[p + α \cos (θ + \frac{2 π}{3})]}^{2} + {[q + β \sin (θ + \frac{2 π}{3})]}^{2} = 1$ $\Rightarrow {[p - \frac{α}{2} (\cos θ + \sqrt{3} \sin θ)]}^{2} + {[q - \frac{β}{2} (\sin θ - \sqrt{3} \cos θ)]}^{2} = 1$ $\Rightarrow p^{2} - α p (\cos θ + \sqrt{3} \sin θ) + \frac{α^{2}}{4} {(\cos θ + \sqrt{3} \sin θ)}^{2} + q^{2} - β q (\sin θ - \sqrt{3} \cos θ) + \frac{β^{2}}{4} {(\sin θ - \sqrt{3} \cos θ)}^{2} = 1 . . . (i i)$ $x_{C} = a [p + α \cos (θ - \frac{2 π}{3})]$ $y_{C} = b [q + β \sin (θ - \frac{2 π}{3})]$ $\Rightarrow {[p + α \cos (θ - \frac{2 π}{3})]}^{2} + {[q + β \sin (θ - \frac{2 π}{3})]}^{2} = 1$ $\Rightarrow {[p - \frac{α}{2} (\cos θ - \sqrt{3} \sin θ)]}^{2} + {[q - \frac{β}{2} (\sin θ + \sqrt{3} \cos θ)]}^{2} = 1$ $\Rightarrow p^{2} - α p (\cos θ - \sqrt{3} \sin θ) + \frac{α^{2}}{4} {(\cos θ - \sqrt{3} \sin θ)}^{2} + q^{2} - β q (\sin θ + \sqrt{3} \cos θ) + \frac{β^{2}}{4} {(\sin θ + \sqrt{3} \cos θ)}^{2} = 1 . . . (i i i)$ $(i) - (i i) :$ $\Rightarrow α (3 \cos θ + \sqrt{3} \sin θ) p + \frac{1}{4} α^{2} (3 \cos θ + \sqrt{3} \sin θ) (\cos θ - \sqrt{3} \sin θ) + β (3 \sin θ - \sqrt{3} \cos θ) q + \frac{1}{4} β^{2} (3 \sin θ - \sqrt{3} \cos θ) (\sin θ + \sqrt{3} \cos θ) = 0$ $\Rightarrow 4 α (3 \cos θ + \sqrt{3} \sin θ) p + 4 β (3 \sin θ - \sqrt{3} \cos θ) q = - (α^{2} - β^{2}) (3 \cos 2 θ - \sqrt{3} \sin 2 θ) . . . (I)$ $(i) - (i i i) :$ $\Rightarrow α p (2 \cos θ - \sqrt{3} \sin θ) + \frac{1}{4} α^{2} (3 \cos θ - \sqrt{3} \sin θ) (\cos θ + \sqrt{3} \sin θ) + β q (3 \sin θ + \sqrt{3} \cos θ) + \frac{1}{4} β^{2} (3 \sin θ + \sqrt{3} \cos θ) (\sin θ - \sqrt{3} \cos θ) = 0$ $\Rightarrow 4 α (3 \cos θ - \sqrt{3} \sin θ) p + 4 β (3 \sin θ + \sqrt{3} \cos θ) q = - (α^{2} - β^{2}) (3 \cos 2 θ + \sqrt{3} \sin 2 θ) . . . (I I)$
	Commented by mr W last updated on 06/Jul/19
	$(3cos θ+(√3)sin θ)αp+(3sin θ−(√3)cos θ)βq=−((α^2 −β^2 )/4)(3cos 2θ−(√3)sin 2θ) ...(i) (3cos θ−(√3)sin θ)αp+(3sin θ+(√3)cos θ)βq=−((α^2 −β^2 )/4)(3cos 2θ+(√3)sin 2θ) ...(ii) D=(3cos θ+(√3)sin θ)(3sin θ+(√3)cos θ)−(3cos θ−(√3)sin θ)(3sin θ−(√3)cos θ) =(6sin 2θ+3(√3))−(6sin 2θ−3(√3)) =6(√3) −(4/(α^2 −β^2 ))(αp)D=−(3sin θ−(√3)cos θ)(3cos 2θ+(√3)sin 2θ)+(3sin θ+(√3)cos θ)(3cos 2θ−(√3)sin 2θ) =6(√3)(cos θcos 2θ−sin θsin 2θ) =6(√3)cos 3θ ⇒αp=−((α^2 −β^2 )/4)cos 3θ −(4/(α^2 −β^2 ))(βq)D=−(3cos θ−(√3)sin θ)(3cos 2θ−(√3)sin 2θ)+(3cos θ+(√3)sin θ)(3cos 2θ+(√3)sin 2θ) =6(√3)(sin θcos 2θ+cos θsin 2θ) =6(√3)sin 3θ ⇒βq=−((α^2 −β^2 )/4)sin 3θ put them into (p+α cos θ)^2 +(q+β sin θ)^2 =1 we get b^2 (cos θ+((a^2 −b^2 )/(4b^2 ))cos 3θ)^2 +a^2 (sin θ+((a^2 −b^2 )/(4a^2 ))sin 3θ)^2 =((3a^2 b^2 )/s^2 ) b^2 cos^2 θ[1+(((a^2 −b^2 ))/b^2 )(cos^2 θ−(3/4))]^2 +a^2 sin^2 θ[1+(((a^2 −b^2 ))/a^2 )((3/4)−sin^2 θ)]^2 =((3a^2 b^2 )/s^2 ) b^2 (cos^2 θ−(1/2)+(1/2))[1+(((a^2 −b^2 ))/b^2 )(cos^2 θ−(1/2)−(1/4))]^2 +a^2 ((1/2)−cos^2 θ+(1/2))[1+(((a^2 −b^2 ))/a^2 )(cos^2 θ−(1/2)+(1/4))]^2 =((3a^2 b^2 )/s^2 ) let Φ=cos^2 θ−(1/2) b^2 (Φ+(1/2))[1+(((a^2 −b^2 ))/b^2 )(Φ−(1/4))]^2 +a^2 ((1/2)−Φ)[1+(((a^2 −b^2 ))/a^2 )(Φ+(1/4))]^2 =((3a^2 b^2 )/s^2 ) a^2 (2Φ+1)[4b^2 +(a^2 −b^2 )(4Φ−1)]^2 +b^2 (1−2Φ)[4a^2 +(a^2 −b^2 )(4Φ+1)]^2 =((96a^4 b^4 )/s^2 ) 32(a^2 −b^2 )^3 Φ^3 −6(a^2 −b^2 )^3 Φ+(a^2 +b^2 )[(a^2 +b^2 )^2 +12a^2 b^2 ]−((96a^4 b^4 )/s^2 )=0 [(a^2 −b^2 )Φ]^3 −(3/(16))(a^2 −b^2 )^2 [(a^2 −b^2 )Φ]+{(1/(32))(a^2 +b^2 )[(a^2 +b^2 )^2 +12a^2 b^2 ]−((3a^4 b^4 )/s^2 )}=0 with μ=(b/a)<1, ξ=(s/a) [(1−μ^2 )Φ]^3 −(3/(16))(1−μ^2 )^2 [(1−μ^2 )Φ]+{(1/(32))(1+μ^2 )[(1+μ^2 )^2 +12μ^2 ]−((3μ^4 )/ξ^2 )}=0 with z=(1−μ^2 )Φ=(1−μ^2 )(cos^2 θ−(1/2)) u=(1/(16))(1−μ^2 )^2 v=(1/(64))(1+μ^2 )[(1+μ^2 )^2 +12μ^2 ]−((3μ^4 )/(2ξ^2 )) ⇒z^3 −3uz+2v=0 Δ=−u^3 +v^2 <0 z=2(√u) sin ((1/3)sin^(−1) (v/(√u^3 ))+((2nπ)/3)) ⇒θ=cos^(−1) (√((1/2)+((2(√μ))/(1−μ^2 )) sin ((1/3)sin^(−1) (v/(√u^3 ))+((2nπ)/3))))$
	$(3 \cos θ + \sqrt{3} \sin θ) α p + (3 \sin θ - \sqrt{3} \cos θ) β q = - \frac{α^{2} - β^{2}}{4} (3 \cos 2 θ - \sqrt{3} \sin 2 θ) . . . (i)$ $(3 \cos θ - \sqrt{3} \sin θ) α p + (3 \sin θ + \sqrt{3} \cos θ) β q = - \frac{α^{2} - β^{2}}{4} (3 \cos 2 θ + \sqrt{3} \sin 2 θ) . . . (i i)$ $D = (3 \cos θ + \sqrt{3} \sin θ) (3 \sin θ + \sqrt{3} \cos θ) - (3 \cos θ - \sqrt{3} \sin θ) (3 \sin θ - \sqrt{3} \cos θ)$ $= (6 \sin 2 θ + 3 \sqrt{3}) - (6 \sin 2 θ - 3 \sqrt{3})$ $= 6 \sqrt{3}$ $- \frac{4}{α^{2} - β^{2}} (α p) D = - (3 \sin θ - \sqrt{3} \cos θ) (3 \cos 2 θ + \sqrt{3} \sin 2 θ) + (3 \sin θ + \sqrt{3} \cos θ) (3 \cos 2 θ - \sqrt{3} \sin 2 θ)$ $= 6 \sqrt{3} (\cos θ \cos 2 θ - \sin θ \sin 2 θ)$ $= 6 \sqrt{3} \cos 3 θ$ $\Rightarrow α p = - \frac{α^{2} - β^{2}}{4} \cos 3 θ$ $- \frac{4}{α^{2} - β^{2}} (β q) D = - (3 \cos θ - \sqrt{3} \sin θ) (3 \cos 2 θ - \sqrt{3} \sin 2 θ) + (3 \cos θ + \sqrt{3} \sin θ) (3 \cos 2 θ + \sqrt{3} \sin 2 θ)$ $= 6 \sqrt{3} (\sin θ \cos 2 θ + \cos θ \sin 2 θ)$ $= 6 \sqrt{3} \sin 3 θ$ $\Rightarrow β q = - \frac{α^{2} - β^{2}}{4} \sin 3 θ$ $p u t t h e m i n t o$ ${(p + α \cos θ)}^{2} + {(q + β \sin θ)}^{2} = 1$ $w e g e t$ $b^{2} {(\cos θ + \frac{a^{2} - b^{2}}{4 b^{2}} \cos 3 θ)}^{2} + a^{2} {(\sin θ + \frac{a^{2} - b^{2}}{4 a^{2}} \sin 3 θ)}^{2} = \frac{3 a^{2} b^{2}}{s^{2}}$ $b^{2} \cos^{2} θ {[1 + \frac{(a^{2} - b^{2})}{b^{2}} (\cos^{2} θ - \frac{3}{4})]}^{2} + a^{2} \sin^{2} θ {[1 + \frac{(a^{2} - b^{2})}{a^{2}} (\frac{3}{4} - \sin^{2} θ)]}^{2} = \frac{3 a^{2} b^{2}}{s^{2}}$ $b^{2} (\cos^{2} θ - \frac{1}{2} + \frac{1}{2}) {[1 + \frac{(a^{2} - b^{2})}{b^{2}} (\cos^{2} θ - \frac{1}{2} - \frac{1}{4})]}^{2} + a^{2} (\frac{1}{2} - \cos^{2} θ + \frac{1}{2}) {[1 + \frac{(a^{2} - b^{2})}{a^{2}} (\cos^{2} θ - \frac{1}{2} + \frac{1}{4})]}^{2} = \frac{3 a^{2} b^{2}}{s^{2}}$ $l e t Φ = \cos^{2} θ - \frac{1}{2}$ $b^{2} (Φ + \frac{1}{2}) {[1 + \frac{(a^{2} - b^{2})}{b^{2}} (Φ - \frac{1}{4})]}^{2} + a^{2} (\frac{1}{2} - Φ) {[1 + \frac{(a^{2} - b^{2})}{a^{2}} (Φ + \frac{1}{4})]}^{2} = \frac{3 a^{2} b^{2}}{s^{2}}$ $a^{2} (2 Φ + 1) {[4 b^{2} + (a^{2} - b^{2}) (4 Φ - 1)]}^{2} + b^{2} (1 - 2 Φ) {[4 a^{2} + (a^{2} - b^{2}) (4 Φ + 1)]}^{2} = \frac{96 a^{4} b^{4}}{s^{2}}$ $32 {(a^{2} - b^{2})}^{3} Φ^{3} - 6 {(a^{2} - b^{2})}^{3} Φ + (a^{2} + b^{2}) [{(a^{2} + b^{2})}^{2} + 12 a^{2} b^{2}] - \frac{96 a^{4} b^{4}}{s^{2}} = 0$ ${[(a^{2} - b^{2}) Φ]}^{3} - \frac{3}{16} {(a^{2} - b^{2})}^{2} [(a^{2} - b^{2}) Φ] + {\frac{1}{32} (a^{2} + b^{2}) [{(a^{2} + b^{2})}^{2} + 12 a^{2} b^{2}] - \frac{3 a^{4} b^{4}}{s^{2}}} = 0$ $w i t h μ = \frac{b}{a} < 1, ξ = \frac{s}{a}$ ${[(1 - μ^{2}) Φ]}^{3} - \frac{3}{16} {(1 - μ^{2})}^{2} [(1 - μ^{2}) Φ] + {\frac{1}{32} (1 + μ^{2}) [{(1 + μ^{2})}^{2} + 12 μ^{2}] - \frac{3 μ^{4}}{ξ^{2}}} = 0$ $w i t h$ $z = (1 - μ^{2}) Φ = (1 - μ^{2}) (\cos^{2} θ - \frac{1}{2})$ $u = \frac{1}{16} {(1 - μ^{2})}^{2}$ $v = \frac{1}{64} (1 + μ^{2}) [{(1 + μ^{2})}^{2} + 12 μ^{2}] - \frac{3 μ^{4}}{2 ξ^{2}}$ $\Rightarrow z^{3} - 3 u z + 2 v = 0$ $Δ = - u^{3} + v^{2} < 0$ $z = 2 \sqrt{u} \sin (\frac{1}{3} \sin^{- 1} \frac{v}{\sqrt{u^{3}}} + \frac{2 n π}{3})$ $\Rightarrow θ = \cos^{- 1} \sqrt{\frac{1}{2} + \frac{2 \sqrt{μ}}{1 - μ^{2}} \sin (\frac{1}{3} \sin^{- 1} \frac{v}{\sqrt{u^{3}}} + \frac{2 n π}{3})}$
	Commented by ajfour last updated on 03/Jul/19

	$N o w i t s b e a u t y, S i r!$ $D o e s i t i m p l y, o n l y o n e r e a l r o o t f o r$ $\cos^{2} θ, a, b, s b e i n g g i v e n .$
	Commented by ajfour last updated on 02/Jul/19

	$y o u r f a i t h i s b e y o n d a l l r e a l m s!$
	Commented by ajfour last updated on 02/Jul/19

	$P l e a s e e x p l a i n w h a t i s D a n d$ $h o w y o u e x p r e s s i t, S i r . .$
	Commented by mr W last updated on 02/Jul/19

	$t h a n k y o u f o r f o l l o w i n g s i r!$ $w i t h D i j u s t {d o n}^{'} t w a n t t o w r i t e$ $t o o m u c h w h e n s o l v i n g t h e e q u a t i o n$ $s y s t e m .$
	Commented by mr W last updated on 02/Jul/19

	Commented by mr W last updated on 03/Jul/19

	$\frac{3 μ^{4}}{ξ^{2}} = {(1 - μ^{2})}^{3} Φ^{3} - \frac{3}{16} {(1 - μ^{2})}^{3} Φ + \frac{1}{32} {(1 + μ^{2})}^{3} + \frac{3}{8} μ^{2} (1 + μ^{2})$ $\frac{d}{d Φ} (\frac{3 μ^{4}}{ξ^{2}}) = 0$ $\Rightarrow 3 Φ^{2} - \frac{3}{16} = 0$ $\Rightarrow Φ = \pm \frac{1}{4} \Rightarrow \cos^{2} θ = \pm \frac{1}{4} + \frac{1}{2} \Rightarrow \cos θ = \pm \frac{\sqrt{3}}{2}, \pm \frac{1}{2}$ $\Rightarrow m a x . s o r m i n . s a t θ = \frac{n π}{2} \pm \frac{π}{6}$ $w i t h Φ = \frac{1}{4} :$ $\frac{3 μ^{4}}{ξ^{2}} = - \frac{1}{32} {(1 - μ^{2})}^{3} + \frac{1}{32} {(1 + μ^{2})}^{3} + \frac{3}{8} μ^{2} (1 + μ^{2})$ $\frac{3 μ^{2}}{ξ^{2}} = \frac{{(3 + μ^{2})}^{2}}{16}$ $\Rightarrow ξ^{2} = \frac{48 μ^{2}}{{(3 + μ^{2})}^{2}}$ $\Rightarrow ξ_{m a x} = \frac{4 \sqrt{3} μ}{3 + μ^{2}} = \frac{s_{m a x}}{a}$ $w i t h Φ = - \frac{1}{4} :$ $\frac{3 μ^{4}}{ξ^{2}} = \frac{1}{32} {(1 - μ^{2})}^{3} + \frac{1}{32} {(1 + μ^{2})}^{3} + \frac{3}{8} μ^{2} (1 + μ^{2})$ $\frac{3 μ^{4}}{ξ^{2}} = \frac{{(1 + 3 μ^{2})}^{2}}{16}$ $\Rightarrow ξ^{2} = \frac{48 μ^{4}}{{(1 + 3 μ^{2})}^{2}}$ $\Rightarrow ξ_{m i n} = \frac{4 \sqrt{3} μ^{2}}{1 + 3 μ^{2}} = \frac{s_{m i n}}{a}$
	Answered by ajfour last updated on 06/Jul/19
	$let mid-point of BC be M(h,k). slope of AM be m=tan θ x_A =h+((s(√3))/2)cos θ, y_A =k+((s(√3))/2)sin θ x_B =h−(s/2)sin θ , y_B =k+(s/2)cos θ x_C =h+(s/2)sin θ , y_C =k−(s/2)cos θ As B,C lie on ellipse (x_B ^2 /a^2 )+(y_B ^2 /b^2 )=1 , (x_C ^2 /a^2 )+(y_C ^2 /b^2 )=1 subtracting (((x_B −x_C )(x_B +x_C ))/a^2 )+(((y_B −y_C )(y_B +y_C ))/b^2 )=0 ⇒ −2b^2 shsin θ+2a^2 skcos θ=0 ⇒ k=((b^2 h)/a^2 )tan θ similarly as A,B lie on ellipse ⇒ ((b^2 s)/2)((√3)cos θ+sin θ)[2h+(s/2)((√3)cos θ−sin θ)] +((a^2 s)/2)((√3)sin θ−cos θ)[2k+(s/2)((√3)sin θ+cos θ)]=0 ⇒ 2hb^2 ((√3)cos θ+sin θ) +2b^2 h(((sin θ)/(cos θ)))((√3)sin θ−cos θ) +(s/2){b^2 (3cos^2 θ−sin^2 θ)+a^2 (3sin^2 θ−cos^2 θ)}=0 ⇒ h=−((scos θ)/(4(√3)b^2 )){a^2 (3sin^2 θ−cos^2 θ)+b^2 (3cos^2 θ−sin^2 θ)} And as A lies on ellipse, b^2 (h+((s(√3))/2)cos θ)^2 +a^2 (((b^2 h)/a^2 )tan θ+((s(√3))/2)sin θ)^2 =a^2 b^2 ⇒ b^2 h^2 +b^2 sh(√3)cos θ+((3b^2 s^2 cos^2 θ)/4) +((b^4 h^2 tan^2 θ)/a^2 )+b^2 sh(√3)sin θtan θ +((3a^2 s^2 sin^2 θ)/4)=a^2 b^2 ⇒ b^2 h^2 (1+(b^2 /a^2 )tan^2 θ)+((b^2 sh(√3))/(cos θ)) +((3s^2 )/4)(b^2 cos^2 θ+a^2 sin^2 θ)=a^2 b^2 ⇒ (b^2 /a^2 )(a^2 cos^2 θ+b^2 sin^2 θ)(h^2 /(cos^2 θ)) +(b^2 s(√3))(h/(cos θ))+(3/4)s^2 (b^2 cos^2 θ+a^2 sin^2 θ) −a^2 b^2 =0 let cos θ=t , then (b^2 /a^2 )((h/t))^2 {b^2 +(a^2 −b^2 )t^2 }+b^2 s(√3)((h/t)) +((3s^2 )/4)[a^2 −(a^2 −b^2 )t^2 ] −a^2 b^2 = 0 where h=−((scos θ)/(4(√3)b^2 )){a^2 (3sin^2 θ−cos^2 θ)+b^2 (3cos^2 θ−sin^2 θ)} ⇒ (h/t)=−(s/(4(√3)b^2 )){3a^2 −b^2 −4(a^2 −b^2 )t^2 } Now (b^2 /a^2 )((s^2 /(48b^4 ))){b^2 +(a^2 −b^2 )t^2 }{3a^2 −b^2 −4(a^2 −b^2 )t^2 }^2 −b^2 s(√3)((s/(4(√3)b^2 ))){3a^2 −b^2 −4(a^2 −b^2 )t^2 } +((3s^2 )/4)[a^2 −(a^2 −b^2 )t^2 ] −a^2 b^2 = 0 let (a^2 −b^2 )t^2 =v (s^2 /(48a^2 b^2 ))(b^2 +v)(3a^2 −b^2 −4v)^2 +(s^2 /4)(b^2 +v)−a^2 b^2 =0 let b^2 +v= y ⇒ ((s^2 /(48a^2 b^2 )))y{3(a^2 +b^2 )−4y}^2 +((s^2 y)/4)−a^2 b^2 = 0 _________________________ ⇒ s^2 y{[3(a^2 +b^2 )−4y]^2 +12a^2 b^2 } −48a^4 b^4 =0 _________________________$
	$l e t m i d - p o i n t o f B C b e M (h, k) .$ $s l o p e o f A M b e m = \tan θ$ $x_{A} = h + \frac{s \sqrt{3}}{2} \cos θ, y_{A} = k + \frac{s \sqrt{3}}{2} \sin θ$ $x_{B} = h - \frac{s}{2} \sin θ, y_{B} = k + \frac{s}{2} \cos θ$ $x_{C} = h + \frac{s}{2} \sin θ, y_{C} = k - \frac{s}{2} \cos θ$ $A s B, C l i e o n e l l i p s e$ $\frac{x_{B}^{2}}{a^{2}} + \frac{y_{B}^{2}}{b^{2}} = 1, \frac{x_{C}^{2}}{a^{2}} + \frac{y_{C}^{2}}{b^{2}} = 1$ $s u b t r a c t i n g$ $\frac{(x_{B} - x_{C}) (x_{B} + x_{C})}{a^{2}} + \frac{(y_{B} - y_{C}) (y_{B} + y_{C})}{b^{2}} = 0$ $\Rightarrow - 2 b^{2} s h \sin θ + 2 a^{2} s k \cos θ = 0$ $\Rightarrow k = \frac{b^{2} h}{a^{2}} \tan θ$ $s i m i l a r l y a s A, B l i e o n e l l i p s e$ $\Rightarrow \frac{b^{2} s}{2} (\sqrt{3} \cos θ + \sin θ) [2 h + \frac{s}{2} (\sqrt{3} \cos θ - \sin θ)]$ $+ \frac{a^{2} s}{2} (\sqrt{3} \sin θ - \cos θ) [2 k + \frac{s}{2} (\sqrt{3} \sin θ + \cos θ)] = 0$ $\Rightarrow$ $2 {h b}^{2} (\sqrt{3} \cos θ + \sin θ)$ $+ 2 b^{2} h (\frac{\sin θ}{\cos θ}) (\sqrt{3} \sin θ - \cos θ)$ $+ \frac{s}{2} {b^{2} (3 \cos^{2} θ - \sin^{2} θ) + a^{2} (3 \sin^{2} θ - \cos^{2} θ)} = 0$ $\Rightarrow h = - \frac{s \cos θ}{4 \sqrt{3} b^{2}} {a^{2} (3 \sin^{2} θ - \cos^{2} θ) + b^{2} (3 \cos^{2} θ - \sin^{2} θ)}$ $A n d a s A l i e s o n e l l i p s e,$ $b^{2} {(h + \frac{s \sqrt{3}}{2} \cos θ)}^{2} + a^{2} {(\frac{b^{2} h}{a^{2}} \tan θ + \frac{s \sqrt{3}}{2} \sin θ)}^{2} = a^{2} b^{2}$ $\Rightarrow b^{2} h^{2} + b^{2} s h \sqrt{3} \cos θ + \frac{3 b^{2} s^{2} \cos^{2} θ}{4}$ $+ \frac{b^{4} h^{2} \tan^{2} θ}{a^{2}} + b^{2} s h \sqrt{3} \sin θ \tan θ$ $+ \frac{3 a^{2} s^{2} \sin^{2} θ}{4} = a^{2} b^{2}$ $\Rightarrow b^{2} h^{2} (1 + \frac{b^{2}}{a^{2}} \tan^{2} θ) + \frac{b^{2} s h \sqrt{3}}{\cos θ}$ $+ \frac{3 s^{2}}{4} (b^{2} \cos^{2} θ + a^{2} \sin^{2} θ) = a^{2} b^{2}$ $\Rightarrow \frac{b^{2}}{a^{2}} (a^{2} \cos^{2} θ + b^{2} \sin^{2} θ) \frac{h^{2}}{\cos^{2} θ}$ $+ (b^{2} s \sqrt{3}) \frac{h}{\cos θ} + \frac{3}{4} s^{2} (b^{2} \cos^{2} θ + a^{2} \sin^{2} θ)$ $- a^{2} b^{2} = 0$ $l e t \cos θ = t, t h e n$ $\frac{b^{2}}{a^{2}} {(\frac{h}{t})}^{2} {b^{2} + (a^{2} - b^{2}) t^{2}} + b^{2} s \sqrt{3} (\frac{h}{t})$ $+ \frac{3 s^{2}}{4} [a^{2} - (a^{2} - b^{2}) t^{2}] - a^{2} b^{2} = 0$ $w h e r e$ $h = - \frac{s \cos θ}{4 \sqrt{3} b^{2}} {a^{2} (3 \sin^{2} θ - \cos^{2} θ) + b^{2} (3 \cos^{2} θ - \sin^{2} θ)}$ $\Rightarrow \frac{h}{t} = - \frac{s}{4 \sqrt{3} b^{2}} {3 a^{2} - b^{2} - 4 (a^{2} - b^{2}) t^{2}}$ $N o w$ $\frac{b^{2}}{a^{2}} (\frac{s^{2}}{48 b^{4}}) {b^{2} + (a^{2} - b^{2}) t^{2}} {3 a^{2} - b^{2} - 4 (a^{2} - b^{2}) t^{2}}^{2}$ $- b^{2} s \sqrt{3} (\frac{s}{4 \sqrt{3} b^{2}}) {3 a^{2} - b^{2} - 4 (a^{2} - b^{2}) t^{2}}$ $+ \frac{3 s^{2}}{4} [a^{2} - (a^{2} - b^{2}) t^{2}] - a^{2} b^{2} = 0$ $l e t (a^{2} - b^{2}) t^{2} = v$ $\frac{s^{2}}{48 a^{2} b^{2}} (b^{2} + v) {(3 a^{2} - b^{2} - 4 v)}^{2}$ $+ \frac{s^{2}}{4} (b^{2} + v) - a^{2} b^{2} = 0$ $l e t b^{2} + v = y$ $\Rightarrow (\frac{s^{2}}{48 a^{2} b^{2}}) y {3 (a^{2} + b^{2}) - 4 y}^{2}$ $+ \frac{s^{2} y}{4} - a^{2} b^{2} = 0$ $_________________________$ $\Rightarrow s^{2} y {{[3 (a^{2} + b^{2}) - 4 y]}^{2} + 12 a^{2} b^{2}}$ $- 48 a^{4} b^{4} = 0$ $_________________________$
	Commented by mr W last updated on 01/Jul/19

	$g r e a t s o l u t i o n s i r!$
	Commented by mr W last updated on 03/Jul/19

	$s i r, p l e a s e c h e c k :$ $s_{0}^{2} = \frac{48 r^{4} a^{2}}{(1 + r^{2}) [{(1 + r^{2})}^{2} + 3]} \Rightarrow s h o u l d b e m i n . s$ $s_{m i n} = \frac{4 \sqrt{3} b}{3 + (\frac{b^{2}}{a^{2}})} \Rightarrow s h o u l d b e m a x . s .$
	Commented by mr W last updated on 03/Jul/19

	$a n o t h e r q u e s t i o n i s :$ $w h y \frac{b}{a} \in (0, \frac{1}{\sqrt{3}}] ? n o t \frac{b}{a} \in (0, 1) ?$ $(a s s u m e b < a)$
	Commented by mr W last updated on 06/Jul/19

	$i f o u n d o u t :$ $m a x . s i s a l w a y s a t θ = 30 ° o r 90 °$ $m i n . s i s a l w a y s a t θ = 0 ° o r 60 °$
	Commented by mr W last updated on 03/Jul/19

	$w i t h r ⩽ \frac{1}{\sqrt{3}} s o m e t h i n g m u s t b e w r o n g .$ $i n f a c t t h e r e i s s o l u t i o n f o r a n y$ $0 < r < 1 .$ $e g . a = 4, b = 3, \Rightarrow r = 0.75 > \frac{1}{\sqrt{3}} :$ $a t θ = 0 ° : s = 5.8344$ $a t θ = 30 ° : s = 5.8004$ $a t θ = 45 ° : s = 5.8172$ $w i t h θ a s m e a n t i n y o u r s o l u t i o n .$
	Commented by mr W last updated on 06/Jul/19

	Commented by mr W last updated on 06/Jul/19

	$t h i s i s t h e θ i n m y s o l u t i o n .$
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