Question-165122

Question Number 165122 by mr W last updated on 26/Jan/22

Commented by mr W last updated on 26/Jan/22

$${three}\:{vertices}\:{of}\:{the}\:{square}\:{lie}\:{on} \\ $$$${the}\:{parabola}.\:{find}\:{the}\:{locus}\:{of}\:{the} \\ $$$${fourth}\:{vertex}\:{inside}\:{the}\:{parabola}. \\ $$

Commented by aleks041103 last updated on 26/Jan/22

$${does}\:{the}\:{square}\:{have}\:{a}\:{defined}\:{side}\:{length} \\ $$

Commented by mr W last updated on 26/Jan/22

$${no},\:{it}\:{mustn}'{t}\:{have}\:{fixed}\:{side}\:{length}. \\ $$

Answered by aleks041103 last updated on 26/Jan/22

The curve is the central branch of 8+92X^2 −18X^4 +X^6 −28Y+116X^2 Y−11X^4 Y+30Y^2 +19X^2 Y^2 −9Y^3 =0 where X=(x/a) and Y=(y/a)

$${The}\:{curve}\:{is}\:{the}\:{central}\:{branch}\:{of} \\ $$$$\mathrm{8}+\mathrm{92}{X}^{\mathrm{2}} −\mathrm{18}{X}^{\mathrm{4}} +{X}^{\mathrm{6}} −\mathrm{28}{Y}+\mathrm{116}{X}^{\mathrm{2}} {Y}−\mathrm{11}{X}^{\mathrm{4}} {Y}+\mathrm{30}{Y}^{\mathrm{2}} +\mathrm{19}{X}^{\mathrm{2}} {Y}^{\mathrm{2}} −\mathrm{9}{Y}^{\mathrm{3}} =\mathrm{0} \\ $$$${where}\:{X}=\frac{{x}}{{a}}\:{and}\:{Y}=\frac{{y}}{{a}} \\ $$

Commented by aleks041103 last updated on 26/Jan/22

Commented by aleks041103 last updated on 26/Jan/22

$${No}\:{matter}\:{what}\:{I}\:{do},\:{I}\:{can}'{t}\:{separate} \\ $$$${the}\:{three}\:{branches}… \\ $$

Commented by aleks041103 last updated on 26/Jan/22

Commented by mr W last updated on 27/Jan/22

$${great}\:{work}\:{sir}! \\ $$$${i}\:{think}\:{the}\:{complete}\:{locus}\:{has}\:{indeed} \\ $$$${three}\:{branches},\:{since}\:{the}\:{fourth} \\ $$$${vertex}\:{can}\:{be}\:{inside}\:{of}\:{the}\:{parabola} \\ $$$${as}\:{well}\:{as}\:{outside}\:{of}\:{the}\:{parabola}. \\ $$

Commented by mr W last updated on 27/Jan/22