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Question Number 117295 by mohammad17 last updated on 10/Oct/20

Answered by mr W last updated on 10/Oct/20

$$y=r\sin \theta$$$$x=r\cos \theta$$$$r=1+\cos \theta$$$$\frac{dy}{d\theta }=\frac{dr}{d\theta }\sin \theta +r\cos \theta =-{\sin }^2\theta +(1+\cos \theta )\cos \theta$$$$=\cos \theta +\cos 2\theta$$$$\frac{dx}{d\theta }=\frac{dr}{d\theta }\cos \theta -r\sin \theta =-\sin \theta \cos \theta -(1+\cos \theta )\sin \theta$$$$=-\sin \theta -\sin 2\theta$$$$\frac{dy}{dx}=-\frac{\cos \theta +\cos 2\theta }{\sin \theta +\sin 2\theta }=0$$$$\Rightarrow \cos \theta +\cos 2\theta =0$$$$\Rightarrow 2{\cos }^2\theta +\cos \theta -1=0$$$$\Rightarrow (2\cos \theta -1)(\cos \theta +1)=0$$$$\Rightarrow \cos \theta =\frac{1}{2}or-1\left(rejected\right)$$$$\Rightarrow x=(1+\frac{1}{2})\frac{1}{2}=\frac{3}{4}$$$$\Rightarrow y=(1+\frac{1}{2})(\pm \frac{\sqrt{3}}{2})=\pm \frac{3\sqrt{3}}{4}$$$$wegettwopointsatwhichthe$$$$tangentisparalleltoxaxis:$$$$pointA(\frac{3}{4},\frac{3\sqrt{3}}{4})$$$$pointB(\frac{3}{4},-\frac{3\sqrt{3}}{4})$$

Commented by mr W last updated on 10/Oct/20

Commented by mohammad17 last updated on 10/Oct/20

$$tbankyouverymuchsir$$

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