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Question Number 191076 by Mingma last updated on 17/Apr/23

Answered by mr W last updated on 18/Apr/23

$$touchingpointP(p,\frac{1}{p^2})$$$$\tan \theta =-\frac{dy}{dx}=\frac{2}{p^3}$$$$r=p-r\sin \theta =p-\frac{2r}{\sqrt{4+p^6}}$$$$\Rightarrow r=\frac{p}{1+\frac{2}{\sqrt{4+p^6}}}$$$$r=\frac{1}{p^2}-r\cos \theta =\frac{1}{p^2}-\frac{{rp}^3}{\sqrt{4+p^6}}$$$$\Rightarrow r=\frac{1}{p^2(1+\frac{p^3}{\sqrt{4+p^6}})}$$$$\frac{p}{1+\frac{2}{\sqrt{4+p^6}}}=\frac{1}{p^2(1+\frac{p^3}{\sqrt{4+p^6}})}$$$$lett=p^3$$$$1+\frac{2}{\sqrt{4+t^2}}=t(1+\frac{t}{\sqrt{4+t^2}})$$$$2-t^2=(t-1)\sqrt{4+t^2}$$$$2t^2-9t+8=0$$$$\Rightarrow t=\frac{9-\sqrt{17}}{4}=p^3$$$$\Rightarrow p=\sqrt[3]{\frac{9-\sqrt{17}}{4}}$$$$\Rightarrow r=\frac{\sqrt[3]{\frac{9-\sqrt{17}}{4}}}{1+\frac{2}{\sqrt{4+{\left(\frac{9-\sqrt{17}}{4}\right)}^2}}}\approx 0.57626$$

Commented by mehdee42 last updated on 18/Apr/23

$$verynice$$

Commented by mr W last updated on 18/Apr/23

Commented by Mingma last updated on 18/Apr/23

Excellent, sir!

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