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Question Number 44436 by ajfour last updated on 29/Sep/18

Commented by ajfour last updated on 29/Sep/18

$$FindRintermsofr.$$

Answered by MrW3 last updated on 29/Sep/18

Commented by MrW3 last updated on 29/Sep/18

$$assumeeqn.ofparabolaisy=\frac{x^2}{a}$$$$y_C=x_C^2/a=r+r\sin \alpha =r(1+\sin \alpha )$$$$\tan \alpha =\frac{x_C}{2y_C}=\frac{{ax}_C^2}{2{ay}_C\times x_C}=\frac{a}{2x_C}=\frac{\sqrt{a}}{2\sqrt{r(1+\sin \alpha )}}$$$$4{\tan }^2\alpha =\frac{a}{r(1+\sin \alpha )}$$$$4r{\sin }^2\alpha (1+\sin \alpha )=a(1-{\sin }^2\alpha )$$$$4{\sin }^3\alpha +(4+\frac{a}{r}){\sin }^2\alpha -\frac{a}{r}=0$$$$orwith\mu =\frac{a}{r}$$$$4{\sin }^3\alpha +(4+\mu ){\sin }^2\alpha -\mu =0...\left(i\right)$$$$$$$$\Rightarrow \alpha =...intermofr$$$$x_B=x_C+r\cos \alpha =\sqrt{ar(1+\sin \alpha )}+r\cos \alpha$$$$$$$$similarly:$$$$x_E=\sqrt{aR(1+\sin \beta )}+R\cos \beta$$$$4{\sin }^3\beta +(4+\frac{a}{R}){\sin }^2\beta -\frac{a}{R}=0$$$$orwith\lambda =\frac{a}{R}$$$$4{\sin }^3\beta +(4+\lambda ){\sin }^2\beta -\lambda =0...\left(ii\right)$$$$$$$${(R-r)}^2+{(x_E-x_B)}^2={(R+r)}^2$$$$x_E-x_B=2\sqrt{Rr}$$$$\sqrt{aR(1+\sin \beta )}+R\cos \beta -\sqrt{ar(1+\sin \alpha )}-r\cos \alpha =2\sqrt{Rr}$$$$\sqrt{aR(1+\sin \beta )}+R\cos \beta -2\sqrt{Rr}=\sqrt{ar(1+\sin \alpha )}+r\cos \alpha$$$$\sqrt{\frac{1+\sin \beta }{\lambda }}+\frac{\cos \beta }{\lambda }-\frac{2}{\sqrt{\mu \lambda }}=\sqrt{\frac{1+\sin \alpha }{\mu }}+\frac{\cos \alpha }{\mu }...\left(iii\right)$$$$$$$$threeeqn.withthreeunknowns:\alpha ,\beta ,\lambda$$$$.....$$$$forr=2,IfoundR\approx 8.842$$

Commented by ajfour last updated on 29/Sep/18

$$greatwaysir,verynice,thanks.$$

Commented by ajfour last updated on 29/Sep/18

$$x_D-x_A=2\sqrt{Rr}$$$$forC,orF$$$$x_{F}and{x}_{D}aretwovaluesof\mathit{b}.$$$$let\frac{dy}{dx}=2\mathit{x}=\tan \mathit{\theta }$$$$\mathit{b}-\mathit{x}=\mathit{\rho }\sin \mathit{\theta }$$$$\mathit{y}-\mathit{\rho }=\mathit{\rho }\cos \mathit{\theta }$$$$\mathit{y}={\mathit{x}}^2$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$4\rho (1+\cos \theta )=\tan {}^2\theta$$$$8\rho \cos {}^2\frac{\theta }{2}=\frac{4\tan {}^2\frac{\theta }{2}}{{(1-\tan {}^2\frac{\theta }{2})}^2}$$$$let\tan \frac{\theta }{2}=t,then$$$$\frac{2\rho }{1+t^2}=\frac{t^2}{{(1-t^2)}^2}$$$$let{t}^2=z$$$$\Rightarrow 2\rho {(1-z)}^2=z(1+z)$$$$2\rho z^2-4\rho z+2\rho =z+z^2$$$$(2\rho -1)z^2-(4\rho +1)z+2\rho =0$$$$\Rightarrow z=t^2=\frac{4\rho -1\pm \sqrt{{(4\rho +1)}^2-8\rho (2\rho -1)}}{2(2\rho -1)}$$$$\Rightarrow t^2=\frac{4\rho -1-\sqrt{1+16\rho }}{4\rho -2}$$$$Nowb=\frac{\tan \theta }{2}+\rho \sin \theta$$$$=\frac{t}{1-t^2}+\frac{2\rho t}{1+t^2}$$$$for{b}_1,\rho =r$$$$for{b}_2,\rho =R$$$$\left(intheirexpressionsintermsoft\right)$$$$b_2-b_1=2\sqrt{Rr}$$$$....$$

Commented by MrW3 last updated on 29/Sep/18

$$Findingadirectexpressionfor\theta in$$$$termofR\left(andr\right),{that}^{\prime }sgreatsir!$$

Commented by ajfour last updated on 29/Sep/18

$$ihavechangeditentirelysir,$$$$itsalljustthesame\left(parametric\right).$$$$probably,implicitaswell.$$

Commented by behi83417@gmail.com last updated on 29/Sep/18

$$niceworkdone!thanksmaster$$$$sirmrW3.$$

Commented by behi83417@gmail.com last updated on 29/Sep/18

$$sirAjfour!greatquestion\mathrm{\&}also$$$$niceprof.thanksforyournice$$$$viewpoint.$$

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