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Question Number 45313 by ajfour last updated on 11/Oct/18

Commented by ajfour last updated on 11/Oct/18

$$Ifallfiveareasareequal,and$$$$circlehasunitradius,find$$$$\mathit{a},and\mathit{b}ofellipse.$$

Answered by MrW3 last updated on 12/Oct/18

$$areaofcircle=areaofellipse$$$$R=radiusofcircle=1$$$$areaofeachregion=\frac{\pi R^2}{3}$$$$eqn.ofcircle\left(polarsystem\right):$$$$r=R$$$$eqn.ofellipse\left(polarsystem\right):$$$$\frac{r^2{\cos }^2\theta }{a^2}+\frac{r^2{\sin }^2\theta }{b^2}=1$$$$\Rightarrow r^2=\frac{a^2{b}^2}{a^2{\sin }^2\theta +b^2{\cos }^2\theta }$$$$$$$$righttopintersectionpointis(r_1,{\theta }_1)$$$$r_1=R$$$$\frac{a^2{b}^2}{a^2{\sin }^2{\theta }_1+b^2{\cos }^2{\theta }_1}=R^2$$$$a^2{\sin }^2{\theta }_1+b^2{\cos }^2{\theta }_1={\left(\frac{ab}{R}\right)}^2$$$$a^2{\sin }^2{\theta }_1+b^2(1-{\sin }^2{\theta }_1)={\left(\frac{ab}{R}\right)}^2$$$$\Rightarrow {\sin }^2{\theta }_1=\frac{b^2(a^2-R^2)}{R^2(a^2-b^2)}$$$$\Rightarrow {\cos }^2{\theta }_1=1-\frac{b^2(a^2-R^2)}{R^2(a^2-b^2)}=\frac{a^2(R^2-b^2)}{R^2(a^2-b^2)}$$$$\Rightarrow \tan {\theta }_1=\frac{b}{a}\sqrt{\frac{a^2-R^2}{R^2-b^2}}$$$$\Rightarrow {\theta }_1={\tan }^{-1}\left(\frac{b}{a}\sqrt{\frac{a^2-R^2}{R^2-b^2}}\right)$$$$$$$$areaofrightmostregionis$$$$A_R=2{\int }_0^{{\theta }_1}\frac{1}{2}(r^2-R^2)d\theta$$$$={\int }_0^{{\theta }_1}(\frac{a^2{b}^2}{a^2{\sin }^2\theta +b^2{\cos }^2\theta }-R^2)d\theta$$$$=a^2{b}^2{\int }_0^{{\theta }_1}\frac{d\theta }{{\left(a\sin \theta \right)}^2+{\left(b\cos \theta \right)}^2}-R^2{\theta }_1$$$$=ab{\tan }^{-1}\left(\frac{a\tan {\theta }_1}{b}\right)-R^2{\theta }_1$$$$=ab{\tan }^{-1}\left(\sqrt{\frac{a^2-R^2}{R^2-b^2}}\right)-R^2{\tan }^{-1}\left(\frac{b}{a}\sqrt{\frac{a^2-R^2}{R^2-b^2}}\right)=\frac{\pi R^2}{3}$$$$$$$$let\lambda =\frac{a}{R}\Rightarrow a=\lambda R$$$$since\pi ab=\pi R^2\Rightarrow ab=R^2\Rightarrow b=\frac{R^2}{a}=\frac{R}{\lambda }$$$$\Rightarrow {\tan }^{-1}\lambda -{\tan }^{-1}\frac{1}{\lambda }=\frac{\pi }{3}$$$$\Rightarrow {\tan }^{-1}\lambda -(\frac{\pi }{2}-{\tan }^{-1}\lambda )=\frac{\pi }{3}$$$$\Rightarrow {\tan }^{-1}\lambda =\frac{5\pi }{6}$$$$\Rightarrow \lambda =\tan \frac{5\pi }{6}=2+\sqrt{3}$$$$\Rightarrow a=(2+\sqrt{3})R$$$$\Rightarrow b=\frac{R}{2+\sqrt{3}}=(2-\sqrt{3})R$$$$\Rightarrow {\theta }_1={\tan }^{-1}\left(\frac{1}{\lambda }\right)={\tan }^{-1}(2-\sqrt{3})=15°$$

Commented by MrW3 last updated on 12/Oct/18

Commented by ajfour last updated on 12/Oct/18

$$Great;thankyouverymuchSir.$$

Commented by behi83417@gmail.com last updated on 12/Oct/18

$$niceproblemandbeautifulsolution.$$$$thanksalotsirAjfour\mathrm{\&}sirmrW3.$$$$$$

Commented by ajfour last updated on 12/Oct/18

$$canwenowhaveaquestionfrom$$$$you,behiSir.$$

Commented by behi83417@gmail.com last updated on 12/Oct/18

$$\text{You can't use 'macro parameter character \#' in math mode}$$$$andwaitingforyourattention.$$

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