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Question Number 131420 by ajfour last updated on 04/Feb/21

Commented by ajfour last updated on 04/Feb/21

$$Intermsofellipseparameters$$$$aandb,findradiusofthe$$$$equalradiicircles.$$

Answered by mr W last updated on 04/Feb/21

Commented by mr W last updated on 05/Feb/21

$$let\mu =\frac{b}{a}$$$$sayP(a+a\cos \varphi ,b+b\sin \varphi )$$$$\tan \theta =\frac{b}{a\tan \varphi }=\frac{\mu }{\tan \varphi }$$$$\tan \varphi =\frac{\mu }{\tan \theta }$$$$\sin \varphi =\frac{\mu }{\sqrt{{\mu }^2+{\tan }^2\theta }}$$$$\cos \varphi =\frac{\tan \theta }{\sqrt{{\mu }^2+{\tan }^2\theta }}$$$$y=b+b\sin \varphi -\frac{\mu }{\tan \varphi }(x-a-a\cos \varphi )$$$$0=b(1+\sin \varphi )-\frac{\mu }{\tan \varphi }(x_A-a(1+\cos \varphi ))$$$$(1+\sin \varphi )=\frac{1}{\tan \varphi }(\frac{x_A}{a}-(1+\cos \varphi ))$$$$\Rightarrow \frac{x_A}{a}=1+\cos \varphi +(1+\sin \varphi )\tan \varphi$$$$\Rightarrow \frac{x_A}{a}=1+\frac{\tan \theta }{\sqrt{{\mu }^2+{\tan }^2\theta }}+(1+\frac{\mu }{\sqrt{{\mu }^2+{\tan }^2\theta }})\frac{\mu }{\tan \theta }$$$$\Rightarrow \frac{x_A}{a}=1+\frac{\mu }{\tan \theta }+\sqrt{1+{\left(\frac{\mu }{\tan \theta }\right)}^2}$$$$$$$$sayQ(a+a\cos \alpha ,b-b\sin \alpha )$$$$\tan \phi =\frac{\mu }{\tan \alpha }$$$$\tan \alpha =\frac{\mu }{\tan \phi }$$$$\sin \alpha =\frac{\mu }{\sqrt{{\mu }^2+{\tan }^2\phi }}$$$$\cos \alpha =\frac{\tan \phi }{\sqrt{{\mu }^2+{\tan }^2\phi }}$$$$x_C=a+a\cos \alpha +r\sin \phi =x_A-\frac{r}{\tan \frac{\theta }{2}}$$$$1+\cos \alpha +\lambda \sin \phi =1+\frac{\mu }{\tan \theta }+\sqrt{1+{\left(\frac{\mu }{\tan \theta }\right)}^2}-\frac{\lambda }{\tan \frac{\theta }{2}}$$$$let\lambda =\frac{r}{a}$$$$\frac{\tan \phi }{\sqrt{{\mu }^2+{\tan }^2\phi }}+\lambda (\sin \phi +\frac{1}{\tan \frac{\theta }{2}})=\frac{\mu }{\tan \theta }+\sqrt{1+{\left(\frac{\mu }{\tan \theta }\right)}^2}$$$$y_C=b-b\sin \alpha -r\cos \phi =r$$$$\mu (1-\sin \alpha )=\lambda (1+\cos \phi )$$$$\lambda =\frac{r}{a}=\frac{\mu }{1+\cos \phi }(1-\frac{\mu }{\sqrt{{\mu }^2+{\tan }^2\phi }})$$$$\frac{\tan \phi }{\sqrt{{\mu }^2+{\tan }^2\phi }}+\frac{\mu }{1+\cos \phi }(1-\frac{\mu }{\sqrt{{\mu }^2+{\tan }^2\phi }})(\sin \phi +\frac{1}{\tan \frac{\theta }{2}})=\frac{\mu }{\tan \theta }+\sqrt{1+{\left(\frac{\mu }{\tan \theta }\right)}^2}...\left(I\right)$$$$similarly$$$$\frac{r}{b}=\frac{1}{\mu (1+\cos \delta )}(1-\frac{1}{\sqrt{1+{\mu }^2{\tan }^2\delta }})$$$$\frac{\mu \tan \delta }{\sqrt{1+{\mu }^2{\tan }^2\delta }}+\frac{1}{\mu (1+\cos \delta )}(1-\frac{1}{\sqrt{1+{\mu }^2{\tan }^2\delta }})(\sin \delta +\frac{1}{\tan (\frac{\pi }{4}-\frac{\theta }{2})})=\frac{\tan \theta }{\mu }+\sqrt{1+{\left(\frac{\tan \theta }{\mu }\right)}^2}...\left(II\right)$$$$\frac{1}{\mu (1+\cos \delta )}(1-\frac{1}{\sqrt{1+{\mu }^2{\tan }^2\delta }})=\frac{1}{1+\cos \phi }(1-\frac{\mu }{\sqrt{{\mu }^2+{\tan }^2\phi }})...\left(III\right)$$$$threeeqn.forthreeunknowns:\theta ,\phi ,\delta .$$$$......$$

Answered by ajfour last updated on 04/Feb/21

Commented by mr W last updated on 05/Feb/21

$$ifoundnowaytoreducetheproblem$$$$totwoequationswithtwounknowns.$$

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