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Question Number 84689 by mr W last updated on 15/Mar/20

Commented by mr W last updated on 15/Mar/20

$$theparabolaisrolledalongthecircle$$$$asshown.$$$$findtheequationoftheparabolawhen$$$$thetouchingpointisattheposition\theta .$$

Commented by mr W last updated on 16/Mar/20

Answered by mr W last updated on 15/Mar/20

Commented by mr W last updated on 16/Mar/20

$$sayinlocalsystem{x}^{\prime }{y}^{\prime }thepointP$$$$is(p,p^2).$$$$BP=AP=a\theta$$$$BP={\int }_0^p\sqrt{1+{\left(2x\right)}^2}dx=\frac{{\sinh }^{-1}\left(2p\right)+\left(2p\right)\sqrt{1+{\left(2p\right)}^2}}{4}$$$$\Rightarrow {\sinh }^{-1}\left(2p\right)+\left(2p\right)\sqrt{1+{\left(2p\right)}^2}=4a\theta$$$$\tan \phi =y^{\prime }=2p$$$$\Rightarrow \phi ={\tan }^{-1}\left(2p\right)$$$$$$$$inxy-system:$$$$P(a\sin \theta ,a\cos \theta )$$$$inclinationangleof{x}^{\prime }-axis:$$$$\alpha =\pi -\theta -\phi$$$$x_B=a\sin \theta -p^2\sin \alpha +p\cos \alpha$$$$\Rightarrow x_B=a\sin \theta -p^2\sin (\theta +\phi )-p\cos (\theta +\phi )$$$$y_B=a\cos \theta +p^2\cos \alpha +p\sin \alpha$$$$\Rightarrow y_B=a\cos \theta -p^2\cos (\theta +\phi )+p\sin (\theta +\phi )$$$$x_F=x_B+\frac{1}{4}\sin \alpha$$$$\Rightarrow x_F=a\sin \theta +(\frac{1}{4}-p^2)\sin (\theta +\phi )-p\cos (\theta +\phi )$$$$y_F=y_B-\frac{1}{4}\cos \alpha$$$$\Rightarrow y_F=a\cos \theta +(\frac{1}{4}-p^2)\cos (\theta +\phi )+p\sin (\theta +\phi )$$$$similarly$$$$\Rightarrow x_E=a\sin \theta -(\frac{1}{4}+p^2)\sin (\theta +\phi )-p\cos (\theta +\phi )$$$$\Rightarrow y_E=a\cos \theta -(\frac{1}{4}+p^2)\cos (\theta +\phi )+p\sin (\theta +\phi )$$$$$$$$eqn.ofdirectix:$$$$y-y_E=\tan \alpha (x-x_E)$$$$\tan (\theta +\phi )(x-x_E)+y-y_E=0$$$$$$$$eqn.ofrolledparabolainxy-system:$$$$distancetodirectrix=distancetofocus$$$$\frac{\tan (\theta +\phi )(x-x_E)+y-y_E}{\sqrt{{\tan }^2(\theta +\phi )+1}}=\sqrt{{(x-x_F)}^2+{(y-y_F)}^2}$$$$\Rightarrow {[\tan (\theta +\phi )(x-x_E)+y-y_E]}^2=[1+{\tan }^2(\theta +\phi )][{(x-x_F)}^2+{(y-y_F)}^2]$$

Commented by mr W last updated on 16/Mar/20

Commented by mr W last updated on 16/Mar/20

Commented by mr W last updated on 16/Mar/20

Commented by mr W last updated on 16/Mar/20

Commented by mr W last updated on 16/Mar/20

Commented by mr W last updated on 17/Mar/20

$$findtheparabolawithitsinitial$$$$orientation$$$$$$$$\alpha =\pi -\theta -\phi =-\pi$$$$\Rightarrow \theta =2\pi -\phi =2\pi -{\tan }^{-1}\left(2p\right)$$$$\Rightarrow {\sinh }^{-1}\left(2p\right)+\left(2p\right)\sqrt{1+{\left(2p\right)}^2}=4a[2\pi -{\tan }^{-1}\left(2p\right)]$$

Commented by mr W last updated on 17/Mar/20

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