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Question Number 166481 by ajfour last updated on 20/Feb/22

Commented by ajfour last updated on 20/Feb/22

$$Findtheequationoftheparabola.$$

Answered by MJS_new last updated on 20/Feb/22

$$\mathrm{I}\mathrm{set}2r=1\Rightarrow$$$$\mathrm{par}:y={ax}^2+(1-a)\mathrm{with}a\approx .594384753544$$$$\mathrm{using}\mathrm{the}\mathrm{method}\mathrm{of}\mathrm{common}\mathrm{tangents}\mathrm{in}P$$$$\mathrm{get}\mathrm{an}\mathrm{equation}\mathrm{of}{5}^{\mathrm{th}}\mathrm{degree}\mathrm{for}p=x_P$$$$p^5-2p^4+\frac{25}{4}{p}^3-\frac{7}{4}{p}^2+\frac{3}{4}p-\frac{1}{4}=0$$$$\mathrm{which}\mathrm{I}\mathrm{cannot}\mathrm{solve}\mathrm{exactly}\mathrm{and}$$$$a=\frac{1-2p}{4p\sqrt{p-p^2}}=\frac{1}{p^2+1}-\frac{p}{2(p^2+1)=\sqrt{p-p^2}}$$

Answered by ajfour last updated on 21/Feb/22

$$Letr=1$$$$A(1-\cos \theta ,\sin \theta )$$$$\frac{\cos \theta }{\sin \theta }=2a(1-\cos \theta )$$$$y={ax}^2+2-4a$$$$\sin \theta =a{(1-\cos \theta )}^2+2-4a$$$$\Rightarrow$$$$a=\frac{2-\sin \theta }{4-{(1-\cos \theta )}^2}=\frac{\cos \theta }{2\sin \theta (1-\cos \theta )}$$$$a\approx 0.2972$$

Commented by MJS_new last updated on 21/Feb/22

$$\mathrm{yes},\mathrm{your}r\mathrm{is}2\mathrm{times}\mathrm{my}r\Rightarrow \mathrm{your}a\mathrm{is}\mathrm{half}$$$$\mathrm{of}\mathrm{my}a$$

Commented by ajfour last updated on 21/Feb/22

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