Question Number 194197 by Mingma last updated on 30/Jun/23
Answered by mr W last updated on 01/Jul/23
Commented by mr W last updated on 01/Jul/23
$${x}_{{P}} ={s}\:\mathrm{cos}\:\theta \\ $$$${y}_{{P}} ={s}\:\mathrm{sin}\:\theta \\ $$$${y}_{{P}} ={f}\left({x}_{{P}} \right) \\ $$$${x}_{{Q}} ={s}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({s}\:\mathrm{cos}\:\theta−\sqrt{\mathrm{3}}\:{s}\:\mathrm{sin}\:\theta\right) \\ $$$${x}_{{P}} −\sqrt{\mathrm{3}}\:{y}_{{P}} =\mathrm{2}{x}_{{Q}} \\ $$$${y}_{{Q}} ={s}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}\:{s}\:\mathrm{cos}\:\theta+{s}\:\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\sqrt{\mathrm{3}}\:{x}_{{P}} +{y}_{{P}} =\mathrm{2}{y}_{{Q}} \\ $$$$\Rightarrow{y}_{{P}} =\frac{−\sqrt{\mathrm{3}}\:{x}_{{P}} +{y}_{{P}} }{\mathrm{2}} \\ $$$$\Rightarrow{x}_{{P}} =\frac{{x}_{{Q}} +\sqrt{\mathrm{3}}{y}_{{Q}} }{\mathrm{2}} \\ $$$${locus}\:{of}\:{point}\:{Q}: \\ $$$$\frac{−\sqrt{\mathrm{3}}\:{x}_{{P}} +{y}_{{P}} }{\mathrm{2}}={f}\left(\frac{{x}_{{Q}} +\sqrt{\mathrm{3}}{y}_{{Q}} }{\mathrm{2}}\right) \\ $$$${or} \\ $$$$\frac{−\sqrt{\mathrm{3}}\:{x}+{y}}{\mathrm{2}}={f}\left(\frac{{x}+\sqrt{\mathrm{3}}{y}}{\mathrm{2}}\right) \\ $$$$ \\ $$$${if}\:{locus}\:{of}\:{P}\:{is}\:{a}\:{segment}\:{y}={ax}+{b}, \\ $$$${then}\:{the}\:{locus}\:{of}\:{Q}\:{is} \\ $$$$\frac{−\sqrt{\mathrm{3}}\:{x}+{y}}{\mathrm{2}}={a}×\frac{{x}+\sqrt{\mathrm{3}}{y}}{\mathrm{2}}+{b} \\ $$$$\Rightarrow\left({a}+\sqrt{\mathrm{3}}\right){x}+\left({a}\sqrt{\mathrm{3}}−\mathrm{1}\right){y}+\mathrm{2}{b}=\mathrm{0} \\ $$$${this}\:{is}\:{also}\:{a}\:{segment}.\: \\ $$$$\left({same}\:{segment}\:{rotated}\:{by}\:\mathrm{60}°\:{about}\:{O}\right) \\ $$$$ \\ $$$${similarly}\:{we}\:{can}\:{see}:\:{if}\:{the}\:{locus}\:{of} \\ $$$${P}\:{is}\:{a}\:{circle},\:{then}\:{the}\:{locus}\:{of}\:{Q}\:{is} \\ $$$${also}\:{a}\:{circle}. \\ $$$$\left({same}\:{circle}\:{rotated}\:{by}\:\mathrm{60}°\:{about}\:{O}\right) \\ $$
Commented by mr W last updated on 01/Jul/23
Commented by mr W last updated on 01/Jul/23
Commented by mr W last updated on 01/Jul/23
$${generally}: \\ $$$${locus}\:{of}\:{point}\:{Q}\:{is}\:{the}\:{locus}\:{of}\:{point} \\ $$$${P}\:{rotated}\:{by}\:\mathrm{60}°\:{about}\:{point}\:{O}. \\ $$
Commented by Mingma last updated on 02/Jul/23
Perfect solutions, sir��