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Question Number 130795 by mr W last updated on 29/Jan/21

Commented by mr W last updated on 29/Jan/21

$$findthelocusofpointP.$$

Answered by mr W last updated on 29/Jan/21

Commented by mr W last updated on 29/Jan/21

$$\mu =\frac{b}{a}$$$$sayA(a\cos \theta ,b\sin \theta )$$$$\tan \phi =\frac{b\cos \theta }{a\sin \theta }=\frac{\mu }{\tan \theta }$$$$\Rightarrow \phi ={\tan }^{-1}\left(\frac{\mu }{\tan \theta }\right)$$$$x_P=a\cos \theta -a\cos \theta \cos 2\phi +b(1+\sin \theta )\sin 2\phi$$$$\Rightarrow x_P=a[\cos \theta (1-\cos 2\phi )+\mu (1+\sin \theta )\sin 2\phi ]$$$$y_P=b\sin \theta +a\cos \theta \sin 2\phi +b(1+\sin \theta )\cos 2\phi$$$$\Rightarrow y_P=a[\mu \sin \theta +\cos \theta \sin 2\phi +\mu (1+\sin \theta )\cos 2\phi ]$$

Commented by mr W last updated on 29/Jan/21

Commented by mr W last updated on 29/Jan/21

Commented by mr W last updated on 29/Jan/21

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