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Question Number 45448 by MrW3 last updated on 13/Oct/18

Commented by MrW3 last updated on 13/Oct/18

$$Whenanellipseisrotatedbyan$$$$angle\alpha ,thefivepartialareasareequal.$$$$Findtheequationoftheellipse$$$$for\alpha =90°and45°respectively.$$

Commented by ajfour last updated on 13/Oct/18

Commented by ajfour last updated on 13/Oct/18

$$eq.ofblueellipse:with\beta =\frac{\alpha }{2}$$$$\frac{r^2\cos {}^2(\theta -\beta )}{a^2}+\frac{r^2\sin {}^2(\theta -\beta )}{b^2}=1$$$$\Rightarrow r^2=\frac{a^2{b}^2}{b^2\cos {}^2(\theta -\beta )+a^2\sin {}^2(\theta -\beta )}$$$$$$$$fortheredone$$$$\frac{r^2\cos {}^2(\theta +\beta )}{a^2}+\frac{r^2\sin {}^2(\theta +\beta )}{b^2}=1$$$$Areaofblueellipsein{I}^{st}quadrant$$$$A_b=\frac{5\pi ab}{12}$$$$Areaofredellipsein{I}^{st}quadrant$$$$A_r=\frac{\pi ab}{12}$$$$A_b={\int }_0^{\pi /2}\left(\frac{r^2}{2}d\theta \right)=\frac{5\pi ab}{12}$$$$\Rightarrow \frac{ab}{2}{\tan }^{-1}\left[\frac{a\tan (\theta -\beta )}{b}\right]{\mid }_0^{\pi /2}=\frac{5\pi ab}{12}$$$$\Rightarrow {\tan }^{-1}\left(\frac{a}{b\tan \beta }\right)+{\tan }^{-1}\left(\frac{a\tan \beta }{b}\right)=\frac{5\pi }{6}$$$$\iff \varphi +\psi =\frac{5\pi }{6},andletm=\tan \beta$$$$\tan (\varphi +\psi )=-\frac{1}{\sqrt{3}}$$$$\Rightarrow \frac{a}{b}(\frac{1}{m}+m)=-\frac{1}{\sqrt{3}}(1-\frac{a^2}{b^2})$$$$\Rightarrow {\left(\frac{a}{b}\right)}^2-\sqrt{3}\left(\frac{1}{\sin \beta \cos \beta }\right)\left(\frac{a}{b}\right)-1=0$$$$\Rightarrow \frac{a}{b}=\frac{\sqrt{3}}{\sin \alpha }+\sqrt{\frac{3}{\sin {}^2\alpha }+1}.$$

Commented by ajfour last updated on 13/Oct/18

$$thanksforconfirmingit,Sir.$$

Commented by MrW3 last updated on 13/Oct/18

$$niceandcorrectsir!$$

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