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Question Number 88272 by ajfour last updated on 09/Apr/20

Commented by ajfour last updated on 09/Apr/20

$$Findradiusofsemicirclein$$$$termsofellipseparameters$$$$aandb.(a>b)$$

Answered by mr W last updated on 09/Apr/20

Commented by mr W last updated on 10/Apr/20

$$let\mu =\frac{b}{a}$$$$sayP(a\cos \theta ,b\sin \theta )$$$$\frac{dy}{dx}=\frac{b\cos \theta }{-a\sin \theta }=-\frac{\mu }{\tan \theta }$$$$\tan \phi =-\frac{1}{\frac{dy}{dx}}=\frac{\tan \theta }{\mu }=\frac{\sin \theta }{\mu \cos \theta }$$$$\Rightarrow \sin \phi =\frac{\sin \theta }{\sqrt{{\sin }^2\theta +{\mu }^2{\cos }^2\theta }}$$$$\Rightarrow \cos \phi =\frac{\mu \cos \theta }{\sqrt{{\sin }^2\theta +{\mu }^2{\cos }^2\theta }}$$$$$$$$x_s=a\cos \theta -r\cos \phi$$$$y_s=b\sin \theta -r\sin \phi$$$${(a-x_s)}^2+y_s^2=r^2$$$${(a-a\cos \theta +r\cos \phi )}^2+{(b\sin \theta -r\sin \phi )}^2=r^2$$$$a^2{(1-\cos \theta )}^2+2ar(1-\cos \theta )\cos \phi +b^2{\sin }^2\theta -2br\sin \theta \sin \phi =0$$$$a^2{(1-\cos \theta )}^2+2ar\frac{(1-\cos \theta )\mu \cos \theta }{\sqrt{{\sin }^2\theta +{\mu }^2{\cos }^2\theta }}+b^2{\sin }^2\theta -2br\frac{\sin \theta \sin \theta }{\sqrt{{\sin }^2\theta +{\mu }^2{\cos }^2\theta }}=0$$$$with\lambda =\frac{r}{a}$$$$\Rightarrow {(1-\cos \theta )}^2+{\mu }^2{\sin }^2\theta =2\mu \lambda \frac{1-\cos \theta }{\sqrt{{\sin }^2\theta +{\mu }^2{\cos }^2\theta }}$$$$\Rightarrow \lambda =\frac{[1+{\mu }^2-(1-{\mu }^2)\cos \theta ]\sqrt{{\sin }^2\theta +{\mu }^2{\cos }^2\theta }}{2\mu }...\left(i\right)$$$$$$$$x_R=2x_s-a$$$$x_R=-a(1-2\cos \theta )-2r\cos \phi =-a(1-2\cos \theta )-\frac{2\mu r\cos \theta }{\sqrt{{\sin }^2\theta +{\mu }^2{\cos }^2\theta }}$$$$y_R=2y_s$$$$y_R=2(b\sin \theta -r\sin \phi )=2\sin \theta (b-\frac{r}{\sqrt{{\sin }^2\theta +{\mu }^2{\cos }^2\theta }})$$$$\frac{x_R^2}{a^2}+\frac{y_R^2}{b^2}=1$$$$b^2{[a(1-2\cos \theta )+\frac{2\mu r\cos \theta }{\sqrt{{\sin }^2\theta +{\mu }^2{\cos }^2\theta }}]}^2+4a^2{\sin }^2\theta {(b-\frac{r}{\sqrt{{\sin }^2\theta +{\mu }^2{\cos }^2\theta }})}^2=a^2{b}^2$$$$\Rightarrow {[1-2\cos \theta +\frac{2\mu \lambda \cos \theta }{\sqrt{{\sin }^2\theta +{\mu }^2{\cos }^2\theta }}]}^2+4{\sin }^2\theta {(1-\frac{\lambda }{\mu \sqrt{{\sin }^2\theta +{\mu }^2{\cos }^2\theta }})}^2=1$$$$\Rightarrow (1-{\mu }^2)[{\mu }^4{\cos }^2\theta (1+\cos \theta )+{(1-\cos \theta )}^3]=2{\mu }^4\cos \theta ...\left(ii\right)$$$$$$$$from\left(i\right)and\left(ii\right)wecanget\theta and\lambda .$$$$$$$$example:a=4,b=3\Rightarrow \mu =\frac{3}{4}$$$$\Rightarrow \theta =73.7398°$$$$\Rightarrow \lambda =0.9434$$

Commented by ajfour last updated on 10/Apr/20

$$Thankssir!$$

Commented by liki last updated on 09/Apr/20

$$...\mathit{s}\mathit{o}\mathit{r}\mathit{y}\mathit{m}\mathit{r}\mathit{W},howtogetyourcontactineedhelpforyou;mycontact+255745266946$$

Commented by mr W last updated on 09/Apr/20

$$sorrysir,ificouldhelpyou,icandoit$$$$onlywithinthisforum.so,whenyou$$$$haveaquestion,justpostithere.$$

Commented by liki last updated on 09/Apr/20

$$..thereissomethingiwanttoaskyousir,thuswhyiwrotemywhatssapno$$

Commented by Ar Brandon last updated on 09/Apr/20

$$Ohmy!$$

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