Question Number 185151 by mathlove last updated on 17/Jan/23
Answered by mr W last updated on 20/Jan/23
$${say}\:{the}\:{touching}\:{point}\:{with}\:{the} \\ $$$${parabola}\:{is}\:\left(−{p},\frac{{p}^{\mathrm{2}} }{\mathrm{2}}\right). \\ $$$${the}\:{center}\:{of}\:{circle}\:{is}\:\left(−{R},{h}\right). \\ $$$$\mathrm{tan}\:\theta={p} \\ $$$$−{R}=−{p}+{R}\:\mathrm{sin}\:\theta=−{p}+\frac{{Rp}}{\:\sqrt{{p}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\Rightarrow{R}=\frac{{p}}{\mathrm{1}+\frac{{p}}{\:\sqrt{{p}^{\mathrm{2}} +\mathrm{1}}}} \\ $$$${h}=\frac{{p}^{\mathrm{2}} }{\mathrm{2}}+{R}\:\mathrm{cos}\:\theta=\frac{{p}^{\mathrm{2}} }{\mathrm{2}}+\frac{{R}}{\:\sqrt{{p}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\Rightarrow{h}=\frac{{p}^{\mathrm{2}} }{\mathrm{2}}+\frac{{p}}{{p}+\sqrt{{p}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${eqn}.\:{of}\:{tangent}\:{line}: \\ $$$${y}=\mathrm{2}−\left({x}−\mathrm{4}\right)\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${x}+\sqrt{\mathrm{3}}{y}−\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{4}=\mathrm{0} \\ $$$$\frac{\left(−{R}+\sqrt{\mathrm{3}}{h}−\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{4}\right)}{\:\sqrt{\mathrm{1}^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }}=−{R} \\ $$$$\sqrt{\mathrm{3}}{h}+{R}−\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{4}=\mathrm{0} \\ $$$$\sqrt{\mathrm{3}}\left(\frac{{p}^{\mathrm{2}} }{\mathrm{2}}+\frac{{p}}{{p}+\sqrt{{p}^{\mathrm{2}} +\mathrm{1}}}\right)+\left(\frac{{p}}{\mathrm{1}+\frac{{p}}{\:\sqrt{{p}^{\mathrm{2}} +\mathrm{1}}}}\right)−\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{p}\approx\mathrm{2}.\mathrm{4837} \\ $$$$\Rightarrow{R}\approx\mathrm{1}.\mathrm{2885} \\ $$
Commented by mr W last updated on 20/Jan/23
