asymptote of Lam\'e's cubicAsymptoteOfLamesCubic2008-03-14 19:17:232008-03-16 12:23:41ExampleasymptoteLam\'e's cubic% this is the default PlanetMath preamble. as your knowledge
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We will show that the {\em Lam\'e's cubic}
\begin{align}
x^3+y^3 = a^3,
\end{align}
where $a$ is a positive \PMlinkescapetext{constant}, has the line
$$y = \underbrace{-x}_{g(x)}$$
as its asymptote.\\
Because the equation (1) of the curve is symmetric with respect to $x$ and $y$, the curve is symmetric about the line \,$y = x$.\, From the solved form
\begin{align}
y = \underbrace{\sqrt[3]{a^3-x^3}}_{f(x)}
\end{align}
of (1) we see that every real value of $x$ gives one point of the curve.
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\rput(3.6,-0.2){$x$}
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\rput(1.13,-0.17){$a$}
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\rput(0.5,-4){Lam\'e's cubic\, $y = \sqrt[3]{a^3-x^3}$\, (blue)}
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The difference \,$\Delta = f(x)\!-\!g(x)$\, \PMlinkescapetext{represents} the distance of a point \,$(x,\,y)$\, of the curve and the point of the asserted asymptote\, $y = -x$\, with the same abscissa $x$.\, We multiply the numerator and denominator with the expression \,$(\sqrt[3]{a^3-x^3})^2-x\sqrt[3]{a^3-x^3}+x^2$ for being able to utilise the polynomial formula
$$(u+v)(u^2-uv+v^2) = u^3+v^3,$$
getting
\begin{align*}
\Delta &= f(x)\!-\!g(x)\\
&= \frac{\sqrt[3]{a^3-x^3}+x}{1}\\
&= \frac{(\sqrt[3]{a^3-x^3})^3+x^3}{(\sqrt[3]{a^3-x^3})^2-x\sqrt[3]{a^3-x^3}+x^2}\\
&= \frac{a^3}{(\sqrt[3]{a^3-x^3})^2-x\sqrt[3]{a^3-x^3}+x^2}.
\end{align*}
Thus,\, $\displaystyle \Delta \to \frac{a^3}{\infty+\infty+\infty} = 0$\; when\; $ |x| \to \infty$\; (see the improper limits).\, According to the definition of \PMlinkname{asymptote}{Asymptote}, the line\, $y = -x$\, is asymptote of Lam\'e's cubic.