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Question Number 71739 by Tony Lin last updated on 19/Oct/19

$$findtheasymptoteoffoliumof$$ $$Descartes{x}^3+y^3=3axy,andaisa$$ $$constant>0$$

Answered by Tony Lin last updated on 19/Oct/19

$$lety=tx$$ $$\Rightarrow x^3+{\left(tx\right)}^3=3{atx}^2$$ $$\{\begin{array}{l}x=\frac{3at}{1+t^3}\\ y=\frac{3{at}^2}{1+t^3}\end{array}$$ $$x\to \pm \mathrm{\infty }\Rightarrow t\to -1$$ $$lettheasymptoteL:mx+b$$ $$\underset{t\to -1}{\lim }\frac{y}{x}=-1=m$$ $$\underset{t\to -1}{\lim }[y-(-x)]$$ $$=\underset{x\to -1}{\lim }\frac{3at(t+1)}{(t+1)(t^2-t+1)}$$ $$=-a$$ $$=b$$ $$therefore$$ $$asymptoteL:-x-a$$

Answered by MJS last updated on 19/Oct/19

$$x^3+y^3-3axy=0$$ $$\mathrm{turning}$$ $$x=\frac{\sqrt{2}}{2}(x^{\ast }+y^{\ast })\wedge y=\frac{\sqrt{2}}{2}(x^{\ast }-y^{\ast })$$ $$\frac{3(a+\sqrt{2}{x}^{\ast })}{2}{\left(y^{\ast }\right)}^2-\frac{(3a-\sqrt{2}{x}^{\ast }){\left(x^{\ast }\right)}^2}{2}=0$$ $$\Rightarrow y^{\ast }=\pm \frac{x^{\ast }\sqrt{9a-3\sqrt{2}{x}^{\ast }}}{3\sqrt{a+\sqrt{2}{x}^{\ast }}}$$ $$\mathrm{denominator}\mathrm{not}\mathrm{defined}\mathrm{for}{x}^{\ast }=-\frac{\sqrt{2}a}{2}+0y^{\ast }$$ $$\mathrm{turning}\mathrm{back}$$ $$p=\frac{\sqrt{2}}{2}(x+y)\wedge q=\frac{\sqrt{2}}{2}(x-y)$$ $$\frac{\sqrt{2}}{2}(x+y)=-\frac{\sqrt{2}a}{2}$$ $$y=-x-a\mathrm{is}\mathrm{the}\mathrm{asymptote}$$

Commented byMJS last updated on 19/Oct/19

$$\mathrm{there}\mathrm{might}\mathrm{be}\mathrm{an}\mathrm{easier}\mathrm{path}...$$

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