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Question Number 183344 by mr W last updated on 25/Dec/22

Commented by mr W last updated on 25/Dec/22

$t h e c i r c l e r o l l s a l o n g t h e p a r a b o l a .$ $f i n d t h e l o c u s o f t h e f i x p o i n t P o n$ $t h e c i r c l e .$
	Answered by TheSupreme last updated on 25/Dec/22

	${l e t}^{'} s H t h e i n t e r s e c t i o n b e t w e e n p a r a b l l a a n d c i r c l e$ $s (H - O) = \int_{0}^{x} \sqrt{1 + \frac{4 x^{2}}{a^{2}}} d x$ $a r c s i n (\frac{2 x}{a}) = u$ $\frac{2 x}{a} = s i n h (u)$ $x = \frac{a}{2} s i n h (u)$ $d x = \frac{a}{2} c o s h (u) d u$ $s (H - 0) = \int_{0}^{x} \frac{a}{2} c o s h (u) \sqrt{1 + {s i n h}^{2} (u)} d u =$ $= \int_{0}^{x} \frac{a}{2} {c o s h}^{2} (u) d u = \frac{a}{4} (a r c s i n h (\frac{2 x}{a}) + s i n h (a r c s i n h (\frac{2 x}{a}) c o s h (a r c s i n h (\frac{2 x}{a}))$ $= \frac{a}{4} (\frac{2 x}{a} + \frac{2 x}{a} \sqrt{1 + \frac{4 x^{2}}{a^{2}})} = \frac{x}{2} [1 + \sqrt{1 + \frac{4 x^{2}}{a^{2}}}]$ $θ = \frac{s}{R} = \frac{x}{2 R} [1 + \sqrt{1 + \frac{4 x^{2}}{a^{2}}}]$ $x_{p} = x_{c} - R c o s (θ)$ $y_{p} = y_{c} + R s i n (θ)$ $. . . .$ $x_{c} = x_{H} + R c o s (\emptyset)$ $y_{c} = y_{h} + R s i n (\emptyset)$ $t a n (- \emptyset) = y^{'} = - 2 \frac{x}{a}$ $t a n (\emptyset) = \frac{a}{2 x}$ $c o s (\emptyset) = \frac{1}{\sqrt{1 + \frac{a^{2}}{4 x^{2}}}} = \frac{2 x}{\sqrt{4 x^{2} + a^{2}}}$ $s i n (\emptyset) = \frac{\frac{a}{2 x}}{\sqrt{1 + \frac{a^{2}}{4 x^{2}}}} = \frac{a}{\sqrt{4 x^{2} + a^{2}}}$ $x_{p} = x + \frac{2 x}{\sqrt{4 x^{2} + a^{2}}} - R c o s (\frac{x}{2 R} [1 + \sqrt{1 + \frac{4 x^{2}}{a^{2}}}])$ $y_{p} = - \frac{x^{2}}{a} + \frac{a}{\sqrt{4 x^{2} + a^{2}}} + R s i n (\frac{x}{2 R} [1 + \sqrt{1 + \frac{4 x^{2}}{a^{2}}}])$
	Commented by mr W last updated on 25/Dec/22

	$t h a n k s s i r!$ $w e g o t d i f f e r e n t r e s u l t f o r t h e l e n g t h$ $o f c u r v e .$
	Answered by mr W last updated on 25/Dec/22

	Commented by mr W last updated on 25/Dec/22
	$y=−(x^2 /a) tan θ=−(dy/dx)=((2x)/a) say point Q(q,−(q^2 /a)) tan θ=((2q)/a)=t ⇒q=((at)/2) s=∫_0 ^q (√(1+y′^2 ))dx=∫_0 ^q (√(1+((4x^2 )/a^2 ))) dx s=(a/2)∫_0 ^t (√(1+ξ^2 )) dξ s=a[t(√(1+t^2 ))+ln (t+(√(1+t^2 )))] s=rϕ ⇒ϕ=[t(√(1+t^2 ))+ln (t+(√(1+t^2 )))](a/r) x_C =q+r sin θ=((at)/2)+r sin θ y_C =−(q^2 /a)+r cos θ=−((at^2 )/4)+r cos θ x_P =x_C −ρ sin (θ+ϕ)=((at)/2)+r sin θ−ρ sin (θ+ϕ) y_P =y_C −ρ cos (θ+ϕ)=−((at^2 )/4)+r cos θ−ρ cos (θ+ϕ) let λ=(a/r), μ=(ρ/r) { (((x_P /r)=((λ tan θ)/2)+sin θ−μ sin (θ+ϕ))),(((y_P /r)=−((λ tan^2 θ)/4)+cos θ−μ cos (θ+ϕ))) :} with ϕ=λ[tan θ (√(1+tan^2 θ))+ln (tan θ+(√(1+tan^2 θ)))]$
	$y = - \frac{x^{2}}{a}$ $\tan θ = - \frac{d y}{d x} = \frac{2 x}{a}$ $s a y p o i n t Q (q, - \frac{q^{2}}{a})$ $\tan θ = \frac{2 q}{a} = t$ $\Rightarrow q = \frac{a t}{2}$ $s = \int_{0}^{q} \sqrt{1 + y^{' 2}} d x = \int_{0}^{q} \sqrt{1 + \frac{4 x^{2}}{a^{2}}} d x$ $s = \frac{a}{2} \int_{0}^{t} \sqrt{1 + ξ^{2}} d ξ$ $s = a [t \sqrt{1 + t^{2}} + \ln (t + \sqrt{1 + t^{2}})]$ $s = r φ$ $\Rightarrow φ = [t \sqrt{1 + t^{2}} + \ln (t + \sqrt{1 + t^{2}})] \frac{a}{r}$ $x_{C} = q + r \sin θ = \frac{a t}{2} + r \sin θ$ $y_{C} = - \frac{q^{2}}{a} + r \cos θ = - \frac{{a t}^{2}}{4} + r \cos θ$ $x_{P} = x_{C} - ρ \sin (θ + φ) = \frac{a t}{2} + r \sin θ - ρ \sin (θ + φ)$ $y_{P} = y_{C} - ρ \cos (θ + φ) = - \frac{{a t}^{2}}{4} + r \cos θ - ρ \cos (θ + φ)$ $l e t λ = \frac{a}{r}, μ = \frac{ρ}{r}$ ${\begin{cases} \frac{x_{P}}{r} = \frac{λ \tan θ}{2} + \sin θ - μ \sin (θ + φ) \\ \frac{y_{P}}{r} = - \frac{λ \tan^{2} θ}{4} + \cos θ - μ \cos (θ + φ) \end{cases}$ $w i t h$ $φ = λ [\tan θ \sqrt{1 + \tan^{2} θ} + \ln (\tan θ + \sqrt{1 + \tan^{2} θ})]$
	Commented by mr W last updated on 25/Dec/22

	Commented by mr W last updated on 25/Dec/22

	Commented by mr W last updated on 25/Dec/22

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