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