Asymptotes of an implicit curve

If the slant asymptote exists, we may denote $$k=\lim_{x \to \infty}\dfrac{y}{x}.$$

Then $$0=\lim_{x \to \infty}\left(1+3\cdot \frac{y}{x}-4 \cdot \frac{y^3}{x^3}-\frac{1}{x^2}+\frac{y}{x^3}+\frac{3}{x^3}\right)=1+3k-4k^3=-(k-1)(2k+1)^2.$$

Thus,$$k=1,-\frac{1}{2}.$$

  • If $k=1$, we denote $y=x+b$ and put it into the equation, we have $$-4b^3-12b^2x-9bx^2+b+3=0.$$

Then $$0=\lim_{x \to \infty} \frac{-4b^3-12b^2x-9bx^2+b+3}{x^2}=-9b.$$

Thus,$$b=0.$$

  • If $k=-\dfrac{1}{2}$, we denote $y=-\dfrac{x}{2}+b$ and put it into the equation, we have $$-8b^3+12b^2x+2b-3x+6=0.$$

Then $$0=\lim_{x \to \infty} \dfrac{-8b^3+12b^2x+2b-3x+6}{x}=12b^2-3.$$

Thus, $$b=\pm \frac{1}{2}.$$

As a result, the slant asymptotes are $$y=x, y=-\frac{x}{2}\pm \frac{1}{2}.$$


The question does not give details about how to apply the method of How to find asymptotes of implicit function?, to this particular problem. Here is an application of that method:

In homogeneous coordinates $(X:Y:Z)$, where $(x,y)$ corresponds to $(x:y:1),$ the equation is $$ X^3 + 3X^2Y - 4Y^3 - XZ^2 + YZ^2 + 3Z^3 = 0. \tag1 $$

On the line at infinity, $Z = 0,$ yielding $$ X^3 + 3X^2Y - 4Y^3 = 0, $$ which factors to $$ (X - Y) (X + 2 Y)^2 = 0.$$ This is solved by $X = Y$ or $X = -2Y,$ which says that the asymptotes are of the form $x = y + \text{constant}$ or $x = -2y + \text{constant},$ but this does not say what the constants are.

Continuing with the method of How to find asymptotes of implicit function?, the derivative of the left-hand side of Equation $(1)$ is $$ (3X^2 + 6XY - Z^2)dX + (3X^2 - 12Y^2 + Z^2)dY + (2YZ - 2XZ + 9Z^2) dZ.\tag2 $$

Evaluating $(2)$ at $(X:Y:Z) = (1 : 1 : 0)$ produces $ 9 dX - 9 dY + 0dZ, $ from which we eventually find the asymptote $x = y$ (whose constant term is zero).

But evaluating $(2)$ at $(X:Y:Z) = (-2 : 1 : 0)$ gives only $0 dX + 0 dY + 0 dZ,$ from which we cannot derive the equation of a line. (It is not clear how you got $y = 0$.) So we have to find the constants that give the asymptote lines some other way.

Tags:

Projective Geometry

Calculus

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