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Question Number 52487 by ajfour last updated on 08/Jan/19

Commented by ajfour last updated on 08/Jan/19

$F i n d α_{m a x} a n d α_{m i n} i n t e r m s o f p, q, a, b .$
Commented by mr W last updated on 09/Jan/19

$q u e s t i o n s e e m s t o b e v e r y h a r d .$
Commented by ajfour last updated on 09/Jan/19

$y e s s i r, l e t M j S S i r e n l i g h t e n u s .$
Commented by MJS last updated on 09/Jan/19

$. . . trying . . .$
Commented by MJS last updated on 09/Jan/19

$I {don}^{'} t think we can solve this .$ ${it}^{'} s even hard with given values for a, b, p, q$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19

$i h a v e s o m e c o m m e n t s r e g a r d i n g t h i s q u e s t i o n$ $p l s r e p l y w h o p o s t e d t h e q u e s t i o n . . .$ $1) a r e l e n g t h p a n d q c o n s t a n t v a l u e . . t o f i n d$ $m a x α p o i n t P a n d Q h a v e t o m o v e a w a y a l o n g$ $t h e e l l i p t i c a l p a t h . . . t h e n h o w t h e l e n g t h P R t h a t$ $i s q a n d Q R t h a t i s p r e m a i n s a m e . . .$ $v a l u e o f p a n d q m u s t c h a n g e$ $2) f o r m i n \propto p o i n t P a n d Q c o m e c l o s e r$ $a s a r e s u l t s l e n g t h p s n d q g e t r e d u c e d .$ $f o r m a x α l e n g t h p a n d q g e t i n c r e a s e d .$ $3) w h a t i s t h e b a s i s o f t h i s q u e s t i o n . . .$ $f o r t h e g i v e n f i g u r e c o s α = \frac{{P R}^{2} + {Q R}^{2} - {P Q}^{2}}{2 \times P R \times Q R}$ $c o s α = \frac{p^{2} + q^{2} - {(P Q)}_{c o n s t a n t v a l u e}^{2}}{2 p q}$ $t h e n △ α = 0$ $s o p l s c l a r i f y t h e q u e s t i o n c l e a r l y . . .$
Commented by ajfour last updated on 09/Jan/19

$o k a y M j S S i r, t h a n k s f o r g i v i n g i t a t r y .$
Commented by ajfour last updated on 09/Jan/19

$i^{'} m v e r y s o r r y T a n m a y S i r, q u e s t i o n$ $m i g h t n o t b e c o r r e c t f o r y o u .$
Commented by MJS last updated on 09/Jan/19

$the question is : find all possible triples$ $P Q R for given a, b, p, q$ $first step (hard enough), find all possible$ $points P, R for given a, b, q$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19

$d o t h e q u e s t i n i t s e l f a v a l i d q u e s t i o n . . . i f s o t h e r e$ $m u s t b e c o r e l a t i o n a m o n g p, q a n d α$ $s o m e s t a t e m e n t i s m i s s i n g . . . c o r e l a t i o n$ $a m o n g p a n d q a n d α . . .$
Commented by ajfour last updated on 09/Jan/19

Commented by ajfour last updated on 09/Jan/19
$let R(acos θ, bsin θ) P[acos θ−pcos (θ+φ), bsin θ−psin (θ+φ)] Q(acos θ−qcos φ, bsin θ−qsin φ) let us call cos θ = c_0 , cos φ= c_1 , cos (θ+φ)=c_2 , sin θ=s_0 , sin φ= s_1 , sin (θ+φ)= s_2 . P is on ellipse, so (((ac_0 −qc_1 )^2 )/a^2 )+(((bs_0 −qs_1 )^2 )/b^2 ) = 1 ⇒ ((qc_1 ^2 )/a^2 )+((qs_1 ^2 )/b^2 ) = (2/(ab))(ac_0 c_1 +bs_0 s_1 ) similarly for Q on ellipse, ⇒ ((pc_2 ^2 )/a^2 )+((ps_2 ^2 )/b^2 ) = (2/(ab))(ac_0 c_2 +bs_0 s_2 ) _________________________ rewriting ac_0 c_1 +bs_0 s_1 = (q/(ab))(b^2 c_1 ^2 +a^2 s_1 ^2 )=(q/(ab))h_1 ac_0 c_2 +bs_0 s_2 = (p/(ab))(b^2 c_2 +a^2 s_2 ^2 )=(p/(ab))h_2 from these two eqs. c_0 = ( determinant ((((qh_1 )/(ab)),(bs_1 )),(((ph_2 )/(ab)),(bs_2 )))_ / determinant (((ac_1 ),(bs_1 )),((ac_2 ),(bs_2 )))^ ) , s_0 = ( determinant (((ac_1 ),((qh_1 )/(ab))),((ac_2 ),((ph_2 )/(ab))))_ / determinant (((ac_1 ),(bs_1 )),((ac_2 ),(bs_2 )))^ ) Now c_0 ^2 +s_0 ^2 = 1 ⇒ b^2 (((qh_1 s_2 −ph_2 s_1 )/(s_2 c_1 −c_2 s_1 )))^2 +a^2 (((ph_2 c_1 −qh_1 c_2 )/(s_2 c_1 −c_2 s_1 )))^2 = a^4 b^4 but s_2 c_1 −c_2 s_1 = sin (φ+α)cos φ−cos (φ+α)sin φ = sin α hence b^2 {qsin (φ+α)[b^2 cos^2 φ+a^2 sin^2 φ] −psin φ[b^2 cos^2 (φ+α)+a^2 sin^2 (φ+α)]}^2 +a^2 {pcos φ[b^2 cos^2 (φ+α)+a^2 sin^2 (φ+α)] −qcos (φ+α)[b^2 cos^2 φ+a^2 sin^2 φ]}^2 = a^4 b^4 sin^2 α And with (dα/dφ) = 0, φ can be obtained; hence 𝛂_(max) and 𝛂_(min) .$
$l e t R (a \cos θ, b \sin θ)$ $P [a \cos θ - p \cos (θ + ϕ), b \sin θ - p \sin (θ + ϕ)]$ $Q (a \cos θ - q \cos ϕ, b \sin θ - q \sin ϕ)$ $l e t u s c a l l \cos θ = c_{0}, \cos ϕ = c_{1},$ $\cos (θ + ϕ) = c_{2}, \sin θ = s_{0}, \sin ϕ = s_{1},$ $\sin (θ + ϕ) = s_{2} .$ $P i s o n e l l i p s e, s o$ $\frac{{({a c}_{0} - {q c}_{1})}^{2}}{a^{2}} + \frac{{({b s}_{0} - {q s}_{1})}^{2}}{b^{2}} = 1$ $\Rightarrow \frac{{q c}_{1}^{2}}{a^{2}} + \frac{{q s}_{1}^{2}}{b^{2}} = \frac{2}{a b} ({a c}_{0} c_{1} + {b s}_{0} s_{1})$ $s i m i l a r l y f o r Q o n e l l i p s e,$ $\Rightarrow \frac{{p c}_{2}^{2}}{a^{2}} + \frac{{p s}_{2}^{2}}{b^{2}} = \frac{2}{a b} ({a c}_{0} c_{2} + {b s}_{0} s_{2})$ $_________________________$ $r e w r i t i n g$ ${a c}_{0} c_{1} + {b s}_{0} s_{1} = \frac{q}{a b} (b^{2} c_{1}^{2} + a^{2} s_{1}^{2}) = \frac{q}{a b} h_{1}$ ${a c}_{0} c_{2} + {b s}_{0} s_{2} = \frac{p}{a b} (b^{2} c_{2} + a^{2} s_{2}^{2}) = \frac{p}{a b} h_{2}$ $f r o m t h e s e t w o e q s .$ $c_{0} = \frac{{\| \begin{matrix} \frac{{q h}_{1}}{a b} & {b s}_{1} \\ \frac{{p h}_{2}}{a b} & {b s}_{2} \end{matrix} \|}_{}}{{\| \begin{matrix} {a c}_{1} & {b s}_{1} \\ {a c}_{2} & {b s}_{2} \end{matrix} \|}^{}}, s_{0} = \frac{{\| \begin{matrix} {a c}_{1} & \frac{{q h}_{1}}{a b} \\ {a c}_{2} & \frac{{p h}_{2}}{a b} \end{matrix} \|}_{}}{{\| \begin{matrix} {a c}_{1} & {b s}_{1} \\ {a c}_{2} & {b s}_{2} \end{matrix} \|}^{}}$ $N o w c_{0}^{2} + s_{0}^{2} = 1$ $\Rightarrow b^{2} {(\frac{{q h}_{1} s_{2} - {p h}_{2} s_{1}}{s_{2} c_{1} - c_{2} s_{1}})}^{2} + a^{2} {(\frac{{p h}_{2} c_{1} - {q h}_{1} c_{2}}{s_{2} c_{1} - c_{2} s_{1}})}^{2} = a^{4} b^{4}$ $b u t s_{2} c_{1} - c_{2} s_{1} = \sin (ϕ + α) \cos ϕ - \cos (ϕ + α) \sin ϕ$ $= \sin α$ $h e n c e$ $b^{2} {q \sin (ϕ + α) [b^{2} \cos^{2} ϕ + a^{2} \sin^{2} ϕ]$ ${- p \sin ϕ [b^{2} \cos^{2} (ϕ + α) + a^{2} \sin^{2} (ϕ + α)]}}^{2}$ $+ a^{2} {p \cos ϕ [b^{2} \cos^{2} (ϕ + α) + a^{2} \sin^{2} (ϕ + α)]$ ${- q \cos (ϕ + α) [b^{2} \cos^{2} ϕ + a^{2} \sin^{2} ϕ]}}^{2}$ $= a^{4} b^{4} \sin^{2} α$ $A n d w i t h \frac{d α}{d ϕ} = 0, ϕ c a n b e o b t a i n e d;$ $h e n c e α_{m a x} a n d α_{m i n} .$
Commented by MJS last updated on 09/Jan/19

$I {don}^{'} t think this is necessary$ $we can ‘ ‘ {rotate}^{″} P R around the ellipse .$ $for any chosen R there are (depending on q)$ $0, 1, 2, 3 or 4 possible positions for P$ $(the problem is, we cannot generally solve$ $a polynome of 4^{th} degree)$ $same for Q R (p)$ $try with$ $a = 5 b = 3 (1) p = 7 q = 9 (2) p = 4 q = 6$ $to see where and how the problems occur$ $we need to make out different cases . . .$ $. . . this was to the former comment of Sir$ $Tanmay$
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