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Question Number 30364 by ajfour last updated on 21/Feb/18

Commented by ajfour last updated on 21/Feb/18

$I f t h e p a r a b o l a w i t h f o c u s F_{0}$ $r o l l s o n t h e c i r c u m f e r e n c e o f$ $c i r c l e (c e n t r e d a t o r i g i n a n d$ $h a v i n g r a d i u s r), t h e n f i n d t h e$ $l o c u s o f t h e f o c u s F o f t h e r o l l i n g$ $p a r a b o l a .$
Commented by ajfour last updated on 21/Feb/18

	Answered by mrW2 last updated on 22/Feb/18

	Commented by mrW2 last updated on 23/Feb/18
	$Eqn. of parabola in u−v−system: v=cu^2 with c=(1/(4a)) At point T(u,v): tan φ=(dv/du)=2cu ⇒u=((tan φ)/(2c))=2a tan φ=GT ⇒v=((tan^2 φ)/(4c))=a tan^2 φ=GV Length of parabola VT^(⌢) =l l=∫_0 ^( u) (√(1+((dv/du))^2 )) du=2a∫_0 ^( φ) (√(1+tan^2 φ)) d(tan φ) =a[tan φ (√(1+tan^2 φ))+ln (tan φ+(√(1+tan^2 φ)))] ST^(⌢) =VT^(⌢) =l rθ=a[tan φ (√(1+tan^2 φ))+ln (tan φ+(√(1+tan^2 φ)))] ⇒θ=(a/r)[tan φ (√(1+tan^2 φ))+ln (tan φ+(√(1+tan^2 φ)))] FG=GV−FV=v−a=a(tan^2 φ−1) Point F(x,y) in x−y−system: x=r sin θ+2a tan φ cos (π−φ−θ)−a(tan^2 φ−1) sin (π−φ−θ) ⇒x=r sin θ−2a tan φ cos (φ+θ)−a(tan^2 φ−1) sin (φ+θ) ⇒x(φ)=r sin {(a/r)[tan φ (√(1+tan^2 φ))+ln (tan φ+(√(1+tan^2 φ)))]}−2a tan φ cos {φ+(a/r)[tan φ (√(1+tan^2 φ))+ln (tan φ+(√(1+tan^2 φ)))]}−a(tan^2 φ−1) sin {φ+(a/r)[tan φ (√(1+tan^2 φ))+ln (tan φ+(√(1+tan^2 φ)))]} y=r cos θ+2a tan φ sin (π−φ−θ)+a(tan^2 φ−1) cos (π−φ−θ) y=r cos θ+2a tan φ sin (φ+θ)−a(tan^2 φ−1) cos (φ+θ) ⇒y(φ)=r cos {(a/r)[tan φ (√(1+tan^2 φ))+ln (tan φ+(√(1+tan^2 φ)))]}+2a tan φ sin {φ+(a/r)[tan φ (√(1+tan^2 φ))+ln (tan φ+(√(1+tan^2 φ)))]}−a(tan^2 φ−1) cos {φ+(a/r)[tan φ (√(1+tan^2 φ))+ln (tan φ+(√(1+tan^2 φ)))]} locus of focus F is: x(φ)=r sin {(a/r)[tan φ (√(1+tan^2 φ))+ln (tan φ+(√(1+tan^2 φ)))]}−2a tan φ cos {φ+(a/r)[tan φ (√(1+tan^2 φ))+ln (tan φ+(√(1+tan^2 φ)))]}−a(tan^2 φ−1) sin {φ+(a/r)[tan φ (√(1+tan^2 φ))+ln (tan φ+(√(1+tan^2 φ)))]} y(φ)=r cos {(a/r)[tan φ (√(1+tan^2 φ))+ln (tan φ+(√(1+tan^2 φ)))]}+2a tan φ sin {φ+(a/r)[tan φ (√(1+tan^2 φ))+ln (tan φ+(√(1+tan^2 φ)))]}−a(tan^2 φ−1) cos {φ+(a/r)[tan φ (√(1+tan^2 φ))+ln (tan φ+(√(1+tan^2 φ)))]} with −(π/2)<φ<(π/2)$
	$E q n . o f p a r a b o l a i n u - v - s y s t e m :$ $v = {c u}^{2} w i t h c = \frac{1}{4 a}$ $A t p o i n t T (u, v) :$ $\tan ϕ = \frac{d v}{d u} = 2 c u$ $\Rightarrow u = \frac{\tan ϕ}{2 c} = 2 a \tan ϕ = G T$ $\Rightarrow v = \frac{\tan^{2} ϕ}{4 c} = a \tan^{2} ϕ = G V$ $L e n g t h o f p a r a b o l a \overset{⌢}{V T} = l$ $l = \int_{0}^{u} \sqrt{1 + {(\frac{d v}{d u})}^{2}} d u = 2 a \int_{0}^{ϕ} \sqrt{1 + \tan^{2} ϕ} d (\tan ϕ)$ $= a [\tan ϕ \sqrt{1 + \tan^{2} ϕ} + \ln (\tan ϕ + \sqrt{1 + \tan^{2} ϕ})]$ $\overset{⌢}{S T} = \overset{⌢}{V T} = l$ $r θ = a [\tan ϕ \sqrt{1 + \tan^{2} ϕ} + \ln (\tan ϕ + \sqrt{1 + \tan^{2} ϕ})]$ $\Rightarrow θ = \frac{a}{r} [\tan ϕ \sqrt{1 + \tan^{2} ϕ} + \ln (\tan ϕ + \sqrt{1 + \tan^{2} ϕ})]$ $F G = G V - F V = v - a = a (\tan^{2} ϕ - 1)$ $P o i n t F (x, y) i n x - y - s y s t e m :$ $x = r \sin θ + 2 a \tan ϕ \cos (π - ϕ - θ) - a (\tan^{2} ϕ - 1) \sin (π - ϕ - θ)$ $\Rightarrow x = r \sin θ - 2 a \tan ϕ \cos (ϕ + θ) - a (\tan^{2} ϕ - 1) \sin (ϕ + θ)$ $\Rightarrow x (ϕ) = r \sin {\frac{a}{r} [\tan ϕ \sqrt{1 + \tan^{2} ϕ} + \ln (\tan ϕ + \sqrt{1 + \tan^{2} ϕ})]} - 2 a \tan ϕ \cos {ϕ + \frac{a}{r} [\tan ϕ \sqrt{1 + \tan^{2} ϕ} + \ln (\tan ϕ + \sqrt{1 + \tan^{2} ϕ})]} - a (\tan^{2} ϕ - 1) \sin {ϕ + \frac{a}{r} [\tan ϕ \sqrt{1 + \tan^{2} ϕ} + \ln (\tan ϕ + \sqrt{1 + \tan^{2} ϕ})]}$ $y = r \cos θ + 2 a \tan ϕ \sin (π - ϕ - θ) + a (\tan^{2} ϕ - 1) \cos (π - ϕ - θ)$ $y = r \cos θ + 2 a \tan ϕ \sin (ϕ + θ) - a (\tan^{2} ϕ - 1) \cos (ϕ + θ)$ $\Rightarrow y (ϕ) = r \cos {\frac{a}{r} [\tan ϕ \sqrt{1 + \tan^{2} ϕ} + \ln (\tan ϕ + \sqrt{1 + \tan^{2} ϕ})]} + 2 a \tan ϕ \sin {ϕ + \frac{a}{r} [\tan ϕ \sqrt{1 + \tan^{2} ϕ} + \ln (\tan ϕ + \sqrt{1 + \tan^{2} ϕ})]} - a (\tan^{2} ϕ - 1) \cos {ϕ + \frac{a}{r} [\tan ϕ \sqrt{1 + \tan^{2} ϕ} + \ln (\tan ϕ + \sqrt{1 + \tan^{2} ϕ})]}$ $l o c u s o f f o c u s F i s :$ $x (ϕ) = r \sin {\frac{a}{r} [\tan ϕ \sqrt{1 + \tan^{2} ϕ} + \ln (\tan ϕ + \sqrt{1 + \tan^{2} ϕ})]} - 2 a \tan ϕ \cos {ϕ + \frac{a}{r} [\tan ϕ \sqrt{1 + \tan^{2} ϕ} + \ln (\tan ϕ + \sqrt{1 + \tan^{2} ϕ})]} - a (\tan^{2} ϕ - 1) \sin {ϕ + \frac{a}{r} [\tan ϕ \sqrt{1 + \tan^{2} ϕ} + \ln (\tan ϕ + \sqrt{1 + \tan^{2} ϕ})]}$ $y (ϕ) = r \cos {\frac{a}{r} [\tan ϕ \sqrt{1 + \tan^{2} ϕ} + \ln (\tan ϕ + \sqrt{1 + \tan^{2} ϕ})]} + 2 a \tan ϕ \sin {ϕ + \frac{a}{r} [\tan ϕ \sqrt{1 + \tan^{2} ϕ} + \ln (\tan ϕ + \sqrt{1 + \tan^{2} ϕ})]} - a (\tan^{2} ϕ - 1) \cos {ϕ + \frac{a}{r} [\tan ϕ \sqrt{1 + \tan^{2} ϕ} + \ln (\tan ϕ + \sqrt{1 + \tan^{2} ϕ})]}$ $w i t h - \frac{π}{2} < ϕ < \frac{π}{2}$
	Commented by ajfour last updated on 23/Feb/18

	$T h a n k y o u S i r, y o u h a v e p r e s e n t e d$ $i t a s g o o d a s i t c a n b e!$
	Commented by ajfour last updated on 23/Feb/18

	$T (u, v) u = p$ $v = \frac{u^{2}}{4 a} \Rightarrow \tan ϕ = \frac{u}{2 a} o r \frac{p}{2 a}$ $l = S T = V T$ $l = r θ = \frac{u}{4 a} \sqrt{1 + \frac{u^{2}}{4 a^{2}}} + \frac{1}{2} \ln (u + \sqrt{1 + \frac{u^{2}}{4 a^{2}}})$ $f r o m h e r e w e o b t a i n u i n t e r m s$ $o f θ . A n d v = \frac{u^{2}}{4 a}, ϕ = \tan^{- 1} (\frac{u}{2 a})$ $x_{G} = r \sin θ - p \cos (θ + ϕ)$ $y_{G} = r \cos θ + p \sin (θ + ϕ)$ $l o c u s o f F :$ $x_{F} = r \sin θ - p \cos (θ + ϕ) - (v - a) \sin (θ + ϕ)$ $y_{F} = r \cos θ + u \sin (θ + ϕ) - (v - a) \cos (θ + ϕ) .$
	Commented by mrW2 last updated on 23/Feb/18

	Commented by mrW2 last updated on 23/Feb/18

	Commented by mrW2 last updated on 23/Feb/18

	Commented by mrW2 last updated on 23/Feb/18

	$t h e p i c t u r e s s h o w a n e x a m p l e w i t h$ $r = 4 a n d a = 1$
	Commented by ajfour last updated on 23/Feb/18

	$V e r y g o o d t o s e e t h e i m a g e s S i r .$ $H a d p i c t u r e d t h i s o n l y b y t h e {m i n d}^{'} s$ $e y e .$
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