Question Number 30364 by ajfour last updated on 21/Feb/18
Commented by ajfour last updated on 21/Feb/18
$${If}\:{the}\:{parabola}\:{with}\:{focus}\:{F}_{\mathrm{0}} \\ $$$${rolls}\:{on}\:{the}\:{circumference}\:{of} \\ $$$${circle}\:\left({centred}\:{at}\:{origin}\:{and}\right. \\ $$$$\left.{having}\:{radius}\:{r}\right),\:{then}\:{find}\:{the} \\ $$$${locus}\:{of}\:{the}\:{focus}\:{F}\:{of}\:{the}\:{rolling} \\ $$$${parabola}. \\ $$
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)](Q30503.png)
$${Eqn}.\:{of}\:{parabola}\:{in}\:{u}−{v}−{system}: \\ $$$${v}={cu}^{\mathrm{2}} \:{with}\:{c}=\frac{\mathrm{1}}{\mathrm{4}{a}} \\ $$$$ \\ $$$${At}\:{point}\:{T}\left({u},{v}\right): \\ $$$$\mathrm{tan}\:\phi=\frac{{dv}}{{du}}=\mathrm{2}{cu} \\ $$$$\Rightarrow{u}=\frac{\mathrm{tan}\:\phi}{\mathrm{2}{c}}=\mathrm{2}{a}\:\mathrm{tan}\:\phi={GT} \\ $$$$\Rightarrow{v}=\frac{\mathrm{tan}^{\mathrm{2}} \:\phi}{\mathrm{4}{c}}={a}\:\mathrm{tan}^{\mathrm{2}} \:\phi={GV} \\ $$$${Length}\:{of}\:{parabola}\:\overset{\frown} {{VT}}={l} \\ $$$${l}=\int_{\mathrm{0}} ^{\:{u}} \sqrt{\mathrm{1}+\left(\frac{{dv}}{{du}}\right)^{\mathrm{2}} }\:{du}=\mathrm{2}{a}\int_{\mathrm{0}} ^{\:\phi} \sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}\:{d}\left(\mathrm{tan}\:\phi\right) \\ $$$$={a}\left[\mathrm{tan}\:\phi\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}+\mathrm{ln}\:\left(\mathrm{tan}\:\phi+\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}\right)\right] \\ $$$$\overset{\frown} {{ST}}=\overset{\frown} {{VT}}={l} \\ $$$${r}\theta={a}\left[\mathrm{tan}\:\phi\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}+\mathrm{ln}\:\left(\mathrm{tan}\:\phi+\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}\right)\right] \\ $$$$\Rightarrow\theta=\frac{{a}}{{r}}\left[\mathrm{tan}\:\phi\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}+\mathrm{ln}\:\left(\mathrm{tan}\:\phi+\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}\right)\right] \\ $$$$ \\ $$$${FG}={GV}−{FV}={v}−{a}={a}\left(\mathrm{tan}^{\mathrm{2}} \:\phi−\mathrm{1}\right) \\ $$$$ \\ $$$${Point}\:{F}\left({x},{y}\right)\:{in}\:{x}−{y}−{system}: \\ $$$${x}={r}\:\mathrm{sin}\:\theta+\mathrm{2}{a}\:\mathrm{tan}\:\phi\:\mathrm{cos}\:\left(\pi−\phi−\theta\right)−{a}\left(\mathrm{tan}^{\mathrm{2}} \:\phi−\mathrm{1}\right)\:\mathrm{sin}\:\left(\pi−\phi−\theta\right) \\ $$$$\Rightarrow{x}={r}\:\mathrm{sin}\:\theta−\mathrm{2}{a}\:\mathrm{tan}\:\phi\:\mathrm{cos}\:\left(\phi+\theta\right)−{a}\left(\mathrm{tan}^{\mathrm{2}} \:\phi−\mathrm{1}\right)\:\mathrm{sin}\:\left(\phi+\theta\right) \\ $$$$\Rightarrow{x}\left(\phi\right)={r}\:\mathrm{sin}\:\left\{\frac{{a}}{{r}}\left[\mathrm{tan}\:\phi\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}+\mathrm{ln}\:\left(\mathrm{tan}\:\phi+\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}\right)\right]\right\}−\mathrm{2}{a}\:\mathrm{tan}\:\phi\:\mathrm{cos}\:\left\{\phi+\frac{{a}}{{r}}\left[\mathrm{tan}\:\phi\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}+\mathrm{ln}\:\left(\mathrm{tan}\:\phi+\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}\right)\right]\right\}−{a}\left(\mathrm{tan}^{\mathrm{2}} \:\phi−\mathrm{1}\right)\:\mathrm{sin}\:\left\{\phi+\frac{{a}}{{r}}\left[\mathrm{tan}\:\phi\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}+\mathrm{ln}\:\left(\mathrm{tan}\:\phi+\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}\right)\right]\right\} \\ $$$$ \\ $$$${y}={r}\:\mathrm{cos}\:\theta+\mathrm{2}{a}\:\mathrm{tan}\:\phi\:\mathrm{sin}\:\left(\pi−\phi−\theta\right)+{a}\left(\mathrm{tan}^{\mathrm{2}} \:\phi−\mathrm{1}\right)\:\mathrm{cos}\:\left(\pi−\phi−\theta\right) \\ $$$${y}={r}\:\mathrm{cos}\:\theta+\mathrm{2}{a}\:\mathrm{tan}\:\phi\:\mathrm{sin}\:\left(\phi+\theta\right)−{a}\left(\mathrm{tan}^{\mathrm{2}} \:\phi−\mathrm{1}\right)\:\mathrm{cos}\:\left(\phi+\theta\right) \\ $$$$\Rightarrow{y}\left(\phi\right)={r}\:\mathrm{cos}\:\left\{\frac{{a}}{{r}}\left[\mathrm{tan}\:\phi\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}+\mathrm{ln}\:\left(\mathrm{tan}\:\phi+\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}\right)\right]\right\}+\mathrm{2}{a}\:\mathrm{tan}\:\phi\:\mathrm{sin}\:\left\{\phi+\frac{{a}}{{r}}\left[\mathrm{tan}\:\phi\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}+\mathrm{ln}\:\left(\mathrm{tan}\:\phi+\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}\right)\right]\right\}−{a}\left(\mathrm{tan}^{\mathrm{2}} \:\phi−\mathrm{1}\right)\:\mathrm{cos}\:\left\{\phi+\frac{{a}}{{r}}\left[\mathrm{tan}\:\phi\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}+\mathrm{ln}\:\left(\mathrm{tan}\:\phi+\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}\right)\right]\right\} \\ $$$$ \\ $$$${locus}\:{of}\:{focus}\:{F}\:{is}: \\ $$$${x}\left(\phi\right)={r}\:\mathrm{sin}\:\left\{\frac{{a}}{{r}}\left[\mathrm{tan}\:\phi\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}+\mathrm{ln}\:\left(\mathrm{tan}\:\phi+\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}\right)\right]\right\}−\mathrm{2}{a}\:\mathrm{tan}\:\phi\:\mathrm{cos}\:\left\{\phi+\frac{{a}}{{r}}\left[\mathrm{tan}\:\phi\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}+\mathrm{ln}\:\left(\mathrm{tan}\:\phi+\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}\right)\right]\right\}−{a}\left(\mathrm{tan}^{\mathrm{2}} \:\phi−\mathrm{1}\right)\:\mathrm{sin}\:\left\{\phi+\frac{{a}}{{r}}\left[\mathrm{tan}\:\phi\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}+\mathrm{ln}\:\left(\mathrm{tan}\:\phi+\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}\right)\right]\right\} \\ $$$${y}\left(\phi\right)={r}\:\mathrm{cos}\:\left\{\frac{{a}}{{r}}\left[\mathrm{tan}\:\phi\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}+\mathrm{ln}\:\left(\mathrm{tan}\:\phi+\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}\right)\right]\right\}+\mathrm{2}{a}\:\mathrm{tan}\:\phi\:\mathrm{sin}\:\left\{\phi+\frac{{a}}{{r}}\left[\mathrm{tan}\:\phi\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}+\mathrm{ln}\:\left(\mathrm{tan}\:\phi+\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}\right)\right]\right\}−{a}\left(\mathrm{tan}^{\mathrm{2}} \:\phi−\mathrm{1}\right)\:\mathrm{cos}\:\left\{\phi+\frac{{a}}{{r}}\left[\mathrm{tan}\:\phi\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}+\mathrm{ln}\:\left(\mathrm{tan}\:\phi+\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\phi}\right)\right]\right\} \\ $$$${with}\:−\frac{\pi}{\mathrm{2}}<\phi<\frac{\pi}{\mathrm{2}} \\ $$
Commented by ajfour last updated on 23/Feb/18
$${Thank}\:{you}\:{Sir},\:{you}\:{have}\:{presented} \\ $$$${it}\:{as}\:{good}\:{as}\:{it}\:{can}\:{be}\:! \\ $$
Commented by ajfour last updated on 23/Feb/18
$${T}\left({u},{v}\right)\:\:\:{u}={p} \\ $$$${v}=\frac{{u}^{\mathrm{2}} }{\mathrm{4}{a}}\:\:\:\:\Rightarrow\:\:\:\mathrm{tan}\:\phi=\frac{{u}}{\mathrm{2}{a}}\:\:{or}\:\frac{{p}}{\mathrm{2}{a}} \\ $$$${l}={ST}\:=\:{VT}\: \\ $$$${l}=\:\boldsymbol{{r}\theta}\:=\:\frac{\boldsymbol{{u}}}{\mathrm{4}\boldsymbol{{a}}}\sqrt{\mathrm{1}+\frac{\boldsymbol{{u}}^{\mathrm{2}} }{\mathrm{4}\boldsymbol{{a}}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\boldsymbol{{u}}+\sqrt{\mathrm{1}+\frac{\boldsymbol{{u}}^{\mathrm{2}} }{\mathrm{4}\boldsymbol{{a}}^{\mathrm{2}} }}\:\right) \\ $$$${from}\:{here}\:{we}\:{obtain}\:\boldsymbol{{u}}\:{in}\:{terms} \\ $$$${of}\:\boldsymbol{\theta}\:.\:\:{And}\:\boldsymbol{{v}}=\frac{\boldsymbol{{u}}^{\mathrm{2}} }{\mathrm{4}\boldsymbol{{a}}}\:,\:\:\boldsymbol{\phi}\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\boldsymbol{{u}}}{\mathrm{2}\boldsymbol{{a}}}\right) \\ $$$$ \\ $$$$\boldsymbol{{x}}_{\boldsymbol{{G}}} ={r}\mathrm{sin}\:\theta−{p}\mathrm{cos}\:\left(\theta+\phi\right) \\ $$$$\boldsymbol{{y}}_{\boldsymbol{{G}}} ={r}\mathrm{cos}\:\theta+{p}\mathrm{sin}\:\left(\theta+\phi\right) \\ $$$${locus}\:{of}\:{F}\:: \\ $$$$\boldsymbol{{x}}_{\boldsymbol{{F}}} ={r}\mathrm{sin}\:\theta−{p}\mathrm{cos}\:\left(\theta+\phi\right)−\left({v}−{a}\right)\mathrm{sin}\:\left(\theta+\phi\right) \\ $$$$\boldsymbol{{y}}_{\boldsymbol{{F}}} ={r}\mathrm{cos}\:\theta+{u}\mathrm{sin}\:\left(\theta+\phi\right)−\left({v}−{a}\right)\mathrm{cos}\:\left(\theta+\phi\right)\:. \\ $$
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
$${the}\:{pictures}\:{show}\:{an}\:{example}\:{with} \\ $$$${r}=\mathrm{4}\:{and}\:{a}=\mathrm{1} \\ $$
Commented by ajfour last updated on 23/Feb/18
$${Very}\:{good}\:{to}\:{see}\:{the}\:{images}\:{Sir}. \\ $$$${Had}\:{pictured}\:{this}\:{only}\:{by}\:{the}\:{mind}'{s} \\ $$$${eye}.\: \\ $$