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Question Number 49964 by ajfour last updated on 12/Dec/18

Commented by ajfour last updated on 12/Dec/18

$$Determine\mathit{\alpha }intermsof\mathit{a}and\mathit{b}$$$$ifthecolouredareaisone-fourth$$$$theellipsearea.$$

Commented by MJS last updated on 13/Dec/18

$$\mathrm{turn}\mathrm{it}+90°$$$$\mathrm{let}b=1$$$$y=a\sqrt{1-x^2}$$$$P=\left(\begin{array}{c}-1\\ 0\end{array}\right)Q=\left(\begin{array}{c}q\\ a\sqrt{1-q^2}\end{array}\right)$$$$\mathrm{line}PQ:y=\frac{a\sqrt{1-q^2}}{q+1}(x+1)$$$$\frac{a\sqrt{1-q^2}}{q+1}\underset{-1}{\stackrel{q}{\int }}(x+1)dx+a\underset{q}{\stackrel{1}{\int }}\sqrt{1-x^2}dx=\frac{\pi }{4}a$$$$\Rightarrow \sqrt{1-q^2}=\arcsin q\Rightarrow q\approx .673612\Rightarrow$$$$\alpha =\arctan (.441611\frac{a}{b})$$

Answered by mr W last updated on 13/Dec/18

$$y-b=-\frac{x}{\tan \alpha }=-kx$$$$r\sin \theta -b=-kr\cos \theta$$$$r=\frac{b}{\sin \theta +k\cos \theta }$$$$r^2(\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2})=1$$$$r^2=\frac{a^2{b}^2}{a^2{\sin }^2\theta +b^2{\cos }^2\theta }=\frac{b^2}{{\sin }^2\theta +{\lambda }^2{\cos }^2\theta }$$$$with\lambda =\frac{b}{a}$$$$\frac{b^2}{{(\sin \theta +k\cos \theta )}^2}=\frac{b^2}{{\sin }^2\theta +{\lambda }^2{\cos }^2\theta }$$$$\frac{1}{{(\tan \theta +k)}^2}=\frac{1}{{\tan }^2\theta +{\lambda }^2}$$$$\Rightarrow {\tan }^2\theta +2k\tan \theta +k^2={\tan }^2\theta +{\lambda }^2$$$$\Rightarrow \tan {\theta }_1=\frac{{\lambda }^2-k^2}{2k}$$$$\Rightarrow {\theta }_1={\tan }^{-1}\frac{{\lambda }^2-k^2}{2k}$$$${\int }_{{\theta }_1}^{\frac{\pi }{2}}\frac{1}{2}[\frac{b^2}{{\sin }^2\theta +{\lambda }^2{\cos }^2\theta }-\frac{b^2}{{(\sin \theta +k\cos \theta )}^2}]d\theta =\frac{\pi ab}{4}$$$${\int }_{{\theta }_1}^{\frac{\pi }{2}}[\frac{1}{{\sin }^2\theta +{\lambda }^2{\cos }^2\theta }-\frac{1}{{(\sin \theta +k\cos \theta )}^2}]d\theta =\frac{\pi }{2\lambda }$$$${[\frac{1}{\lambda }{\tan }^{-1}\frac{\tan \theta }{\lambda }+\frac{1}{k+\tan \theta }]}_{{\theta }_1}^{\frac{\pi }{2}}=\frac{\pi }{2\lambda }$$$$[\frac{1}{\lambda }(\frac{\pi }{2}-{\tan }^{-1}\frac{{\lambda }^2-k^2}{2\lambda k})-\frac{1}{k+\frac{{\lambda }^2-k^2}{2k}}]=\frac{\pi }{2\lambda }$$$$[\frac{1}{\lambda }(\frac{\pi }{2}-{\tan }^{-1}\frac{{\lambda }^2-k^2}{2\lambda k})-\frac{2k}{{\lambda }^2+k^2}]=\frac{\pi }{2\lambda }$$$${\tan }^{-1}\frac{k^2-{\lambda }^2}{2\lambda k}=\frac{2\lambda k}{k^2+{\lambda }^2}$$$${\tan }^{-1}\frac{2\lambda k}{k^2-{\lambda }^2}+\frac{2\lambda k}{k^2+{\lambda }^2}=\frac{\pi }{2}$$$$with\mu =\frac{k}{\lambda }$$$${\tan }^{-1}\frac{2}{\mu -\frac{1}{\mu }}+\frac{2}{\mu +\frac{1}{\mu }}=\frac{\pi }{2}$$$$\Rightarrow \mu =\frac{k}{\lambda }=2.2644$$$$\Rightarrow k=\frac{1}{\tan \alpha }=2.2644\lambda =2.2644\frac{b}{a}$$$$\Rightarrow \alpha ={\tan }^{-1}\frac{0.4416a}{b}$$

Commented by mr W last updated on 13/Dec/18

$$sorrysir!Ihadamistake,Ilosta{\tan }^{-1}$$$$sothesolutionbecamesimpleform.$$$$butthisisnottrue.$$

Commented by MJS last updated on 13/Dec/18

$$\mathrm{this}\mathrm{is}\mathrm{different}\mathrm{from}\mathrm{my}\mathrm{solution}...$$$$\mathrm{please}\mathrm{check}\mathrm{mine}$$

Commented by MJS last updated on 13/Dec/18

$$\mathrm{ok}.\mathrm{I}\mathrm{also}\mathrm{went}\mathrm{through}\mathrm{your}\mathrm{solution}\mathrm{but}$$$${\mathrm{couldn}}^{\prime }\mathrm{t}\mathrm{find}\mathrm{a}\mathrm{mistake}.\mathrm{thank}\mathrm{you}\mathrm{for}$$$$\mathrm{checking}!$$

Commented by mr W last updated on 13/Dec/18

$$nowwehavethesameresult.$$

Commented by ajfour last updated on 13/Dec/18

$$Congratulationsandthanks$$$$tobothSirs.$$

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