Question Number 186329 by ajfour last updated on 03/Feb/23
Answered by ajfour last updated on 03/Feb/23
$${Cubic}\:{curve}:\:{y}={x}^{\mathrm{3}} −{x} \\ $$$${Parabola}:\:\:{x}={h}+\left({y}−{k}\right)^{\mathrm{2}} \\ $$$${x}={h}+\left({p}−{k}\right)^{\mathrm{2}} \\ $$$${p}^{\mathrm{3}} −{p}={c} \\ $$$$\Rightarrow\:\:{p}^{\mathrm{2}} =\mathrm{2}{kp}+{x}−{h}−{k}^{\mathrm{2}} \\ $$$${say}\:\:\:\:{x}−{h}−{k}^{\mathrm{2}} ={t} \\ $$$${p}^{\mathrm{2}} =\mathrm{2}{kp}+{t} \\ $$$${p}+{c}=\mathrm{2}{kp}^{\mathrm{2}} +{tp} \\ $$$$\mathrm{2}{kp}^{\mathrm{2}} =\left(\mathrm{1}−{t}\right){p}+{c}=\mathrm{4}{k}^{\mathrm{2}} {p}+\mathrm{2}{kt} \\ $$$$ \\ $$$$\Rightarrow\:\:{p}=\frac{{c}−\mathrm{2}{kt}}{\mathrm{4}{k}^{\mathrm{2}} +{t}−\mathrm{1}}\:\:\:=\frac{{c}−{mt}}{{m}^{\mathrm{2}} +{t}−\mathrm{1}} \\ $$$${now},\:{since}\:\:\:\:{p}^{\mathrm{2}} =\mathrm{2}{kp}+{t} \\ $$$$\left({c}−{mt}\right)^{\mathrm{2}} ={m}\left({c}−{mt}\right)\left({m}^{\mathrm{2}} +{t}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{t}\left({t}+{m}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{c}^{\mathrm{2}} +{m}^{\mathrm{2}} {t}^{\mathrm{2}} −\mathrm{2}{cmt} \\ $$$$={cm}^{\mathrm{3}} +{cmt}−{cm}−{m}^{\mathrm{4}} {t}−{m}^{\mathrm{2}} {t}^{\mathrm{2}} +{m}^{\mathrm{2}} {t} \\ $$$$\:\:\:\:\:\:\:+{t}^{\mathrm{3}} +\mathrm{2}\left({m}^{\mathrm{2}} −\mathrm{1}\right){t}^{\mathrm{2}} +\left({m}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} {t} \\ $$$$\Rightarrow\: \\ $$$${t}^{\mathrm{3}} −\mathrm{2}{t}^{\mathrm{2}} +\left(\mathrm{3}{cm}−{m}^{\mathrm{2}} +\mathrm{1}\right){t} \\ $$$$+{c}\left({m}^{\mathrm{3}} −{m}−{c}\right)=\mathrm{0} \\ $$$$…. \\ $$