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Question Number 71739 by Tony Lin last updated on 19/Oct/19
find the asymptote of folium of Descartes x^3 +y^3 =3axy, and a is a constant >0
$${find}\:{the}\:{asymptote}\:{of}\:{folium}\:{of}\: \\ $$$${Descartes}\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{3}{axy},\:{and}\:{a}\:{is}\:{a} \\ $$$${constant}\:>\mathrm{0} \\ $$
Answered by Tony Lin last updated on 19/Oct/19
let y=tx ⇒x^3 +(tx)^3 =3atx^2 { ((x=((3at)/(1+t^3 )))),((y=((3at^2 )/(1+t^3 )))) :} x→±∞⇒t→−1 let the asymptote L: mx+b lim_(t→−1) (y/x)=−1=m lim_(t→−1) [y−(−x)] =lim_(x→−1) ((3at(t+1))/((t+1)(t^2 −t+1))) =−a =b therefore asymptote L:−x−a
$${let}\:{y}={tx} \\ $$$$\Rightarrow{x}^{\mathrm{3}} +\left({tx}\right)^{\mathrm{3}} =\mathrm{3}{atx}^{\mathrm{2}} \\ $$$$\begin{cases}{{x}=\frac{\mathrm{3}{at}}{\mathrm{1}+{t}^{\mathrm{3}} }}\\{{y}=\frac{\mathrm{3}{at}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{3}} }}\end{cases} \\ $$$${x}\rightarrow\pm\infty\Rightarrow{t}\rightarrow−\mathrm{1} \\ $$$${let}\:{the}\:{asymptote}\:{L}:\:{mx}+{b} \\ $$$$\underset{{t}\rightarrow−\mathrm{1}} {\mathrm{lim}}\frac{{y}}{{x}}=−\mathrm{1}={m} \\ $$$$\underset{{t}\rightarrow−\mathrm{1}} {\mathrm{lim}}\left[{y}−\left(−{x}\right)\right] \\ $$$$=\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{3}{at}\left({t}+\mathrm{1}\right)}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)} \\ $$$$=−{a} \\ $$$$={b} \\ $$$${therefore} \\ $$$${asymptote}\:{L}:−{x}−{a} \\ $$
Answered by MJS last updated on 19/Oct/19
x^3 +y^3 −3axy=0 turning x=((√2)/2)(x^∗ +y^∗ )∧y=((√2)/2)(x^∗ −y^∗ ) ((3(a+(√2)x^∗ ))/2)(y^∗ )^2 −(((3a−(√2)x^∗ )(x^∗ )^2 )/2)=0 ⇒ y^∗ =±((x^∗ (√(9a−3(√2)x^∗ )))/(3(√(a+(√2)x^∗ )))) denominator not defined for x^∗ =−(((√2)a)/2)+0y^∗ turning back p=((√2)/2)(x+y)∧q=((√2)/2)(x−y) ((√2)/2)(x+y)=−(((√2)a)/2) y=−x−a is the asymptote
$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} −\mathrm{3}{axy}=\mathrm{0} \\ $$$$\mathrm{turning} \\ $$$${x}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left({x}^{\ast} +{y}^{\ast} \right)\wedge{y}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left({x}^{\ast} −{y}^{\ast} \right) \\ $$$$\frac{\mathrm{3}\left({a}+\sqrt{\mathrm{2}}{x}^{\ast} \right)}{\mathrm{2}}\left({y}^{\ast} \right)^{\mathrm{2}} −\frac{\left(\mathrm{3}{a}−\sqrt{\mathrm{2}}{x}^{\ast} \right)\left({x}^{\ast} \right)^{\mathrm{2}} }{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\:{y}^{\ast} =\pm\frac{{x}^{\ast} \sqrt{\mathrm{9}{a}−\mathrm{3}\sqrt{\mathrm{2}}{x}^{\ast} }}{\mathrm{3}\sqrt{{a}+\sqrt{\mathrm{2}}{x}^{\ast} }} \\ $$$$\mathrm{denominator}\:\mathrm{not}\:\mathrm{defined}\:\mathrm{for}\:{x}^{\ast} =−\frac{\sqrt{\mathrm{2}}{a}}{\mathrm{2}}+\mathrm{0}{y}^{\ast} \\ $$$$\mathrm{turning}\:\mathrm{back} \\ $$$${p}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left({x}+{y}\right)\wedge{q}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left({x}−{y}\right) \\ $$$$\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left({x}+{y}\right)=−\frac{\sqrt{\mathrm{2}}{a}}{\mathrm{2}} \\ $$$${y}=−{x}−{a}\:\mathrm{is}\:\mathrm{the}\:\mathrm{asymptote} \\ $$
Commented by MJS last updated on 19/Oct/19
there might be an easier path...
$$\mathrm{there}\:\mathrm{might}\:\mathrm{be}\:\mathrm{an}\:\mathrm{easier}\:\mathrm{path}… \\ $$

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