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Question Number 185151 by mathlove last updated on 17/Jan/23

Answered by mr W last updated on 20/Jan/23

$$saythetouchingpointwiththe$$$$parabolais(-p,\frac{p^2}{2}).$$$$thecenterofcircleis(-R,h).$$$$\tan \theta =p$$$$-R=-p+R\sin \theta =-p+\frac{Rp}{\sqrt{p^2+1}}$$$$\Rightarrow R=\frac{p}{1+\frac{p}{\sqrt{p^2+1}}}$$$$h=\frac{p^2}{2}+R\cos \theta =\frac{p^2}{2}+\frac{R}{\sqrt{p^2+1}}$$$$\Rightarrow h=\frac{p^2}{2}+\frac{p}{p+\sqrt{p^2+1}}$$$$eqn.oftangentline:$$$$y=2-(x-4)\frac{1}{\sqrt{3}}$$$$x+\sqrt{3}y-2\sqrt{3}-4=0$$$$\frac{(-R+\sqrt{3}h-2\sqrt{3}-4)}{\sqrt{1^2+{\left(\sqrt{3}\right)}^2}}=-R$$$$\sqrt{3}h+R-2\sqrt{3}-4=0$$$$\sqrt{3}(\frac{p^2}{2}+\frac{p}{p+\sqrt{p^2+1}})+\left(\frac{p}{1+\frac{p}{\sqrt{p^2+1}}}\right)-2\sqrt{3}-4=0$$$$\Rightarrow p\approx 2.4837$$$$\Rightarrow R\approx 1.2885$$

Commented by mr W last updated on 20/Jan/23

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