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Question Number 51134 by ajfour last updated on 24/Dec/18

Commented by ajfour last updated on 24/Dec/18

$$Find\theta intermsofaandbsuchthat$$$$thetwocolouredareasareequal.$$

Answered by mr W last updated on 24/Dec/18

$$\lambda =\frac{a}{b}$$$$r=\frac{ab}{\sqrt{a^2{\sin }^2\alpha +b^2{\cos }^2\alpha }}$$$$r\sin \alpha =(a+r\cos \alpha )\tan \theta$$$$\Rightarrow r(\sin \alpha -\cos \alpha \tan \theta )=a\tan \theta$$$$\Rightarrow \frac{ab(\sin \alpha -\cos \alpha \tan \theta )}{\sqrt{a^2{\sin }^2\alpha +b^2{\cos }^2\alpha }}=a\tan \theta$$$$\Rightarrow \tan \alpha -\tan \theta =\tan \theta \sqrt{{\lambda }^2{\tan }^2\alpha +1}$$$$\Rightarrow {\tan }^2\alpha -2\tan \theta \tan \alpha +{\tan }^2\theta ={\tan }^2\theta ({\lambda }^2{\tan }^2\alpha +1)$$$$\Rightarrow (1-{\lambda }^2{\tan }^2\theta )\tan \alpha =2\tan \theta$$$$\Rightarrow \tan \alpha =\frac{2\tan \theta }{1-{\lambda }^2{\tan }^2\theta }=\frac{2t}{1-{\lambda }^2{t}^2}$$$$A_{yellow}=\frac{1}{2}\times a\tan \theta \times r\cos \alpha +\frac{a^2{b}^2}{2}{\int }_0^{\alpha }\frac{d\alpha }{a^2{\sin }^2\alpha +b^2{\cos }^2\alpha }$$$$A_{yellow}=\frac{a^2\tan \theta }{2\sqrt{{\lambda }^2{\tan }^2\alpha +1}}+\frac{ab}{2}{\tan }^{-1}\left(\frac{a}{b}\tan \alpha \right)=\frac{\pi ab}{8}$$$$\frac{\lambda t(1-{\lambda }^2{t}^2)}{1+{\lambda }^2{t}^2}+{\tan }^{-1}\left(\frac{2\lambda t}{1-{\lambda }^2{\mathrm{t}}^2}\right)=\frac{\pi }{4}$$$$let\mu =\lambda t=\frac{a\tan \theta }{b}$$$$\frac{\mu (1-{\mu }^2)}{1+{\mu }^2}+{\tan }^{-1}\left(\frac{2\mu }{1-{\mu }^2}\right)=\frac{\pi }{4}$$$$\frac{2{\mu }^2}{1+{\mu }^2}\times \frac{1-{\mu }^2}{2\mu }+{\tan }^{-1}\left(\frac{2\mu }{1-{\mu }^2}\right)=\frac{\pi }{4}$$$$let\tan \gamma =\mu =\frac{a\tan \theta }{b}$$$$\Rightarrow \tan 2\gamma =\frac{2\mu }{1-{\mu }^2}$$$$\Rightarrow \frac{{\sin }^2\gamma }{\tan 2\gamma }+\gamma =\frac{\pi }{8}$$$$\Rightarrow \gamma =0.273$$$$\Rightarrow \tan \gamma =0.28$$$$\Rightarrow \tan \theta =\frac{b}{a}\tan \gamma$$$$\Rightarrow \theta ={\tan }^{-1}(0.28\frac{b}{a})$$

Commented by mr W last updated on 24/Dec/18

$$alternative\left(easier\right)way:$$$$incaseofcircle,i.e.b=a=R$$$$A_{yellow}=\theta R^2+\frac{R\tan \theta R\cos 2\theta }{2}=\frac{\pi R^2}{8}$$$$\theta +\frac{\tan \theta \cos 2\theta }{2}=\frac{\pi }{8}$$$$\Rightarrow \theta =0.273$$$$\Rightarrow \tan \theta =0.28$$$$incaseofellipse:$$$$\Rightarrow \tan \theta =0.28\frac{b}{a}$$$$\Rightarrow \theta ={\tan }^{-1}(0.28\frac{b}{a})$$

Commented by ajfour last updated on 24/Dec/18

$$seemedimpossibletome!$$$$\mathbb{SUPERB}Sir!Understood.Thankyou.$$

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