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Question Number 177787 by mr W last updated on 09/Oct/22

Commented by mr W last updated on 09/Oct/22

$$elipsewithparametersaandbis$$$$rotatedaboutitscenterbyangle\phi .$$$$findtheoverlappingareainterms$$$$ofa,b,\phi .$$$$(0⩽\phi ⩽\frac{\pi }{2})$$

Answered by mr W last updated on 09/Oct/22

Commented by mr W last updated on 09/Oct/22

$$redelipse:$$$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$$$\frac{r^2{\cos }^2\theta }{a^2}+\frac{y^2{\sin }^2\theta }{b^2}=1$$$$\Rightarrow r^2=\frac{a^2{b}^2}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }$$$${\theta }_P=\frac{\phi }{2}$$$${\theta }_Q=\frac{\pi +\phi }{2}=\pi -(\frac{\pi }{2}-\frac{\phi }{2})$$$$$$$$areaofsectorOPQofredelipse:$$$$A_{sector}=\frac{1}{2}{\int }_{{\theta }_P}^{{\theta }_Q}{r}^{2}d\theta$$$$overlappingarea=4\times sector$$$$A_{overlapping}=2{\int }_{{\theta }_P}^{{\theta }_Q}{r}^{2}d\theta$$$$=2{\int }_{{\theta }_P}^{{\theta }_Q}\frac{a^2{b}^2}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }d\theta$$$$=2ab{\left[{\tan }^{-1}\left(\frac{a}{b}\times \tan \theta \right)\right]}_{{\theta }_P}^{{\theta }_Q}$$$$=2ab\{\pi -{\tan }^{-1}\left[\frac{a}{b}\times \tan (\frac{\pi }{2}-\frac{\phi }{2})\right]-{\tan }^{-1}\left(\frac{a}{b}\times \tan \frac{\phi }{2}\right)\}$$$$=2ab[\pi -{\tan }^{-1}\left(\frac{a}{b}\cot \frac{\phi }{2}\right)-{\tan }^{-1}\left(\frac{a}{b}\tan \frac{\phi }{2}\right)]$$

Commented by Tawa11 last updated on 09/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

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