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Question Number 136988 by mohammad17 last updated on 28/Mar/21

$$whatstheintersictionofthecurvesr=\frac{1}{1-cos\theta }$$$$$$$$andr=\frac{1}{1+cos\theta }?$$

Answered by Ñï= last updated on 28/Mar/21

$$r=\frac{1}{1-\cos \theta }$$$$\Rightarrow r-r\cos \theta =\sqrt{x^2+y^2}-x=1\left(1\right)$$$$r=\frac{1}{1+\cos \theta }$$$$\Rightarrow r+r\cos \theta =\sqrt{x^2+y^2}+x=1\left(2\right)$$$$\left(2\right)\wedge \left(1\right)$$$$\Rightarrow x=0y=\pm 1$$$$intersectpoints:(0,1),(0,-1)$$

Answered by physicstutes last updated on 29/Mar/21

$$\mathrm{At}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{intersection},\mathrm{these}\mathrm{two}$$$$\mathrm{polar}\mathrm{curves}\mathrm{are}\mathrm{equal}.\mathrm{That}\mathrm{is}$$$$\frac{1}{1-\cos \theta }=\frac{1}{1+\cos \theta }$$$$\Rightarrow 1-\cos \theta =1+\cos \theta$$$$\cos \theta =0\mathrm{or}\theta =2\pi n\pm \frac{\pi }{2}$$$$\mathrm{hence}\theta =\{\frac{\pi }{2},-\frac{\pi }{2}\}$$$$\mathrm{points}\mathrm{of}\mathrm{intersection}\mathrm{are}(1,\frac{\pi }{2})\mathrm{and}(-1,-\frac{\pi }{2})$$$$$$

Commented by physicstutes last updated on 29/Mar/21

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