Question Number 150308 by ajfour last updated on 11/Aug/21
Commented by ajfour last updated on 11/Aug/21
$$\:\:{Q}.\mathrm{149894}\:\left({revisit}\right) \\ $$
Answered by ajfour last updated on 11/Aug/21
$${suppose}\:{s}\:{is}\:{limiting}.\:{Then} \\ $$$${ECP}\:\:{is}\:{a}\:{straight}\:{line}. \\ $$$${let}\:\:\angle{EOA}=\theta.\:{Now} \\ $$$${s}\mathrm{cos}\:\theta+\left({s}\sqrt{\mathrm{3}}\mp{r}\right)\mathrm{sin}\:\theta={h} \\ $$$${s}\mathrm{sin}\:\theta+\left({s}\sqrt{\mathrm{3}}\mp{r}\right)\mathrm{cos}\:\theta={k} \\ $$$$\Rightarrow\: \\ $$$$\frac{\mathrm{cos}\:\theta+\sqrt{\mathrm{3}}\mathrm{sin}\:\theta}{{h}\pm{r}\mathrm{sin}\:\theta}=\frac{\mathrm{sin}\:\theta+\sqrt{\mathrm{3}}\mathrm{cos}\:\theta}{{k}\pm{r}\mathrm{cos}\:\theta} \\ $$$$ \\ $$$$\:{s}_{{m}} =\frac{\pm{r}\left({k}\mathrm{tan}\:\theta−{h}\right)}{{k}−\sqrt{\mathrm{3}}{h}+\left(\sqrt{\mathrm{3}}{k}−{h}\right)\mathrm{tan}\:\theta} \\ $$$$\bigstar \\ $$$${or}\:{from}\:{another}\:{viewpoint} \\ $$$$\left({i}\:{shall}\:{attach}\:{diagram}\right),\:{we} \\ $$$${get} \\ $$$$\frac{\boldsymbol{{r}}}{\boldsymbol{{s}}}=\sqrt{\mathrm{3}}+\boldsymbol{{sin}}\mathrm{2}\boldsymbol{\theta}+\boldsymbol{{cos}}\mathrm{2}\boldsymbol{\theta{tan}}\left(\boldsymbol{\theta}+\boldsymbol{\alpha}\right) \\ $$$$\:\:\:\forall\:\:\boldsymbol{\alpha}=\boldsymbol{{tan}}^{−\mathrm{1}} \left(\frac{\boldsymbol{{k}}}{\boldsymbol{{h}}}\right). \\ $$
Commented by mr W last updated on 11/Aug/21
$${it}\:{seems}\:{not}\:{true}\:{that}\:{ECP}\:{is}\: \\ $$$${straight}\:{line}\:{when}\:{s}\:{is}\:{maximum}\:{or} \\ $$$${minimum}\:{as}\:{shown}\:{by}\:{the}\:{diagram}. \\ $$
Commented by mr W last updated on 11/Aug/21
Commented by ajfour last updated on 11/Aug/21
$${you}\:{could}\:{be}\:{right}\:{sir},\:{i}\:{thought} \\ $$$${it}\:{was}\:{more}\:{probable}\:\left({without}\right. \\ $$$$\left.{proof}\right). \\ $$