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Question Number 50811 by ajfour last updated on 20/Dec/18

Commented by ajfour last updated on 20/Dec/18

$$Finddistancebetweencentres$$$$ofthetwoequalellipses.$$

Commented by ajfour last updated on 21/Dec/18

Answered by mr W last updated on 20/Dec/18

$$y=b\sqrt{1-{\left(\frac{x}{a}\right)}^2}$$$$2{\int }_h^{a}b\sqrt{1-{\left(\frac{x}{a}\right)}^2}dx=\frac{\pi ab}{6}$$$${\int }_h^a\sqrt{1-{\left(\frac{x}{a}\right)}^2}d\left(\frac{x}{a}\right)=\frac{\pi }{12}$$$$\frac{1}{2}{[\left(\frac{x}{a}\right)\sqrt{1-{\left(\frac{x}{a}\right)}^2}+{\sin }^{-1}\left(\frac{x}{a}\right)]}_h^a=\frac{\pi }{12}$$$$\frac{1}{2}[\frac{\pi }{2}-\left(\frac{h}{a}\right)\sqrt{1-{\left(\frac{h}{a}\right)}^2}-{\sin }^{-1}\left(\frac{h}{a}\right)]=\frac{\pi }{12}$$$$\left(\frac{h}{a}\right)\sqrt{1-{\left(\frac{h}{a}\right)}^2}+{\sin }^{-1}\left(\frac{h}{a}\right)=\frac{\pi }{3}$$$$\Rightarrow \frac{h}{a}\approx 0.5533$$$$distancebetweencentersofellipse:$$$$d=2h\approx 1.107a$$

Commented by ajfour last updated on 20/Dec/18

$$BeautifulsolutionSir!Thanks.$$$$\sin 2\theta +2\theta =\frac{2\pi }{3}$$$$d=2a\sin \theta .$$

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