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Question Number 88456 by ajfour last updated on 10/Apr/20

Commented by ajfour last updated on 11/Apr/20

$F i n d s i n t e r m s a a n d b .$
Commented by Tony Lin last updated on 11/Apr/20

Commented by Tony Lin last updated on 11/Apr/20

$k = s c o s θ, h = s s i n θ$ $[\begin{matrix} c o s (- 60 °) & - s i n (- 60 °) \\ s i n (- 60 °) & c o s (- 60 °) \end{matrix}] [\begin{matrix} - k \\ h \end{matrix}]$ $= [\begin{matrix} \frac{\sqrt{3} h - k}{2} \\ \frac{\sqrt{3} k + h}{2} \end{matrix}]$ $P = (\frac{\sqrt{3} h - k}{2} + k, \frac{\sqrt{3} k + h}{2})$ $= (\frac{\sqrt{3} h + k}{2}, \frac{\sqrt{3} k + h}{2})$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $\frac{{(\frac{\sqrt{3} h + k}{2})}^{2}}{a^{2}} + \frac{{(\frac{\sqrt{3} k + h}{2})}^{2}}{b^{2}} = 1$ $k^{2} (b^{2} + 3 a^{2}) + h^{2} (a^{2} + 3 b^{2}) + 2 \sqrt{3} h k (a^{2} + b^{2})$ $s^{2} [{c o s}^{2} θ (b^{2} + 3 a^{2}) + {s i n}^{2} θ (a^{2} + 3 b^{2}) + \sqrt{3} s i n 2 θ (a^{2} + b^{2})$ $= 4 a^{2} b^{2}$ $s = \sqrt{\frac{4 a^{2} b^{2}}{{c o s}^{2} θ (b^{2} + 3 a^{2}) + {s i n}^{2} θ (a^{2} + 3 b^{2}) + \sqrt{3} s i n 2 θ (a^{2} + b^{2})}}$
Commented by ajfour last updated on 11/Apr/20

$T h a n k y o u, T o n y S i r!$ $N i c e s o l u t i o n .$
	Answered by mr W last updated on 11/Apr/20

	Commented by mr W last updated on 11/Apr/20

	$y_{Q} = s \sin θ$ $x_{P} = s \cos (\frac{π}{3} - θ) = \frac{s}{2} (\cos θ + \sqrt{3} \sin θ)$ $y_{P} = s [\sin θ + \sin (\frac{π}{3} - θ)] = \frac{s}{2} (\sin θ + \sqrt{3} \cos θ)$ $b^{2} x_{P}^{2} + a^{2} y_{P}^{2} = a^{2} b^{2}$ $\frac{s^{2}}{4} [b^{2} {(\cos θ + \sqrt{3} \sin θ)}^{2} + a^{2} {(\sin θ + \sqrt{3} \cos θ)}^{2}] = a^{2} b^{2}$ $s^{2} [b^{2} (\cos^{2} θ + 2 \sqrt{3} \sin θ \cos θ + 3 \cos^{2} θ) + a^{2} (\sin^{2} θ + 2 \sqrt{3} \sin θ \cos θ + 3 \sin^{2} θ)] = 4 a^{2} b^{2}$ $s^{2} [\sqrt{3} (a^{2} + b^{2}) \sin 2 θ + (a^{2} + 3 b^{2}) \sin^{2} θ + (3 a^{2} + b^{2}) \cos^{2} θ)] = 4 a^{2} b^{2}$ $w i t h μ = \frac{b}{a}$ $P = \frac{4 b^{2}}{s^{2}} = \sqrt{3} (1 + μ^{2}) \sin 2 θ + (1 + 3 μ^{2}) \sin^{2} θ + (3 + μ^{2}) \cos^{2} θ$ $P = \frac{4 b^{2}}{s^{2}} = 2 (1 + μ^{2}) + \sqrt{3} (1 + μ^{2}) \sin 2 θ + (1 - μ^{2}) \cos 2 θ$ $\frac{d P}{d θ} = 2 \sqrt{3} (1 + μ^{2}) \cos 2 θ - 2 (1 - μ^{2}) \sin 2 θ = 0$ $\Rightarrow \tan 2 θ = \frac{(1 + μ^{2}) \sqrt{3}}{1 - μ^{2}}$ $\Rightarrow \sin 2 θ = \frac{(1 + μ^{2}) \sqrt{3}}{2 \sqrt{1 + μ^{2} + μ^{4}}}$ $\Rightarrow \cos 2 θ = \frac{1 - μ^{2}}{2 \sqrt{1 + μ^{2} + μ^{4}}}$ $\frac{4 b^{2}}{s_{m i n}^{2}} = 2 (1 + μ^{2}) + \sqrt{3} (1 + μ^{2}) \frac{(1 + μ^{2}) \sqrt{3}}{2 \sqrt{1 + μ^{2} + μ^{4}}} + (1 - μ^{2}) \frac{1 - μ^{2}}{2 \sqrt{1 + μ^{2} + μ^{4}}}$ $\frac{2 b^{2}}{s_{m i n}^{2}} = 1 + μ^{2} + \sqrt{1 + μ^{2} + μ^{4}}$ $\Rightarrow s_{m i n} = \frac{\sqrt{2} b}{\sqrt{1 + μ^{2} + \sqrt{1 + μ^{2} + μ^{4}}}}$ $e x a m p l e : a = 4, b = 3, μ = \frac{3}{4}$ $\Rightarrow s_{m i n} = \frac{\sqrt{50 - 2 \sqrt{481}}}{3} \times b = \sqrt{50 - 2 \sqrt{481}} = 2.477$
	Commented by ajfour last updated on 11/Apr/20

	$S i r, h o w d i d y o u k n o w, w i t h d P / d θ = 0,$ $w e g e t s_{m a x} o r s_{m i n} ?$
	Commented by mr W last updated on 11/Apr/20

	Commented by mr W last updated on 11/Apr/20

	$t h e r e a r e i n f i n i t e e q u i l a t e r a l t r i a n g l e s,$ $\frac{d P}{d θ} = 0 f o r m a x . P w h i c h l e a d s t o s_{m i n .}$
	Commented by ajfour last updated on 11/Apr/20

	$S i r, c a n y o u p l e a s e c o m p a r e t h i s$ $i m a g e w i t h t h a t f o r s_{m a x} ?$
	Commented by mr W last updated on 11/Apr/20

	$f r o m$ $P = \frac{4 b^{2}}{s^{2}} = 2 (1 + μ^{2}) + \sqrt{3} (1 + μ^{2}) \sin 2 θ + (1 - μ^{2}) \cos 2 θ$ $w e s e e i t c a n b e f o r m e d t o$ $P = A + B \sin 2 θ + C \cos 2 θ = A + \sqrt{B^{2} + C^{2}} \cos (2 θ - ϕ)$ $i t r e a c h e s m a x . (i . e . s_{m i n}) a t 2 θ = ϕ$ $b u t i t s m i n . (i . e . s_{m a x}) i s a t 2 θ = ϕ \pm π$ $w h i c h l i e s b e h i n d o u r v a l i d r a n g e .$ $t h e r e a l s_{m a x} i s o n t h e b o r d e r o f o u r$ $v a l i d r a n g e .$
	Commented by mr W last updated on 11/Apr/20

	Commented by mr W last updated on 11/Apr/20

	Commented by mr W last updated on 11/Apr/20

	$s i n c e p o i n t Q i s r e s t r i c t e d f r o m (0, 0)$ $t o (0, b) w e {c a n}^{'} t g e t t h e m a t h e m a t i c a l$ $m a x i m u m s_{m a x} . t h e a c t u a l s_{m a x} i s$ $o n e o f t h e t w o c a s e s a b o v e : Q a t (0, 0)$ $o r Q a t (0, b) .$ $b u t s_{m i n} i s b e t w e e n b o t h c a s e s,$ $c o r r e s p o n d s t o \frac{d P}{d θ} = 0 .$
	Commented by ajfour last updated on 11/Apr/20

	$T h a n k y o u S i r!$ $\underline{F o r a = 4, b = 3}$ $t h i s m e a n s f o r θ = 0 w e h a v e$ $s_{m a x} & a t s o m e i n t e r m e d i a t e θ = α,$ $a s s h o w n i n m a i n s o l u t i o n w e$ $h a v e s_{m i n}, a n d θ_{m a x} = β (s a y)$ $α < β < \frac{π}{2} .$ $C o m p l e t e T r e a t m e n t S i r!$ $t h a n k s a l o t a g a i n .$
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