Finding asymptotes for the given curve of two variables

Question

I am looking for the asymptotes for the curve

$x_1=\frac{C}{x_{1}^{2}-3x_1x_2+3x_{2}^{2}}$

where $C\in \mathbb{R}$ and $x_1$ and $x_2$ are the set variables.

Oftentimes the asymptotes are found by factoring out the denominator and setting them to be equal to zero, and in this case in particular we would get

$x_1=\frac{C}{x_{1}^{2}-3x_1x_2+3x_{2}^{2}}$,

thus, is it reasonable say that the asymptotes for the given curve are given by

$-2x_{1}i+(\sqrt{3}+3i)x_{2}=0 \Rightarrow x_{2}=\frac{2x_{1}}{\sqrt{3}+3i}$

and

$2x_{1}i+(\sqrt{3}-3i)x_{2}\Rightarrow x_{2}=\frac{-2x_{1}}{\sqrt{3}-3i}$.

Am I mistaken somewhere or is this the correct way to find the asymptotes for the given curve in the $(x_1,x_2)$-plane?

Answer 1

As

$$ C = x_1(x_1^2-3x_1x_2+3x_2^2) $$

and giving that the real roots of $x_1^2-3x_1x_2+3x_2^2=0$ are null, as $x_1\to 0$ we have $x_2\to \pm \infty$ as long as $C\ne 0$

Answer 2

(see figure below).

With more readable letters, your equation is

$$x(x^{2}-3xy+3y^{2})=C\tag{1}$$

where the factor between parentheses can be written

$$(x-\frac32y)^2+\frac34y^2>0$$

We can assume without loss of generality that $C>0$, because changing $(x,y)$ into its opposite $(-x,-y)$ changes $C$ into its opposite.

First solution with a polar equation:

Setting $x=r \cos \theta, y=r \sin \theta$:

$$r^3 \cos \theta(...) = C \ \ \implies \ \ r=\sqrt[3]{\frac{C}{\cos \theta (...)}}\tag{2}$$

(I haven't given the detailed content of the parentheses because we know that it is always $\ne 0$).

(2) permits to conclude because

$$r \to \infty \iff \theta \to \frac{\pi}{2} or -\frac{\pi}{2}$$

yields a vertical asymptote $x=a$. It remains to check that $a=0$ : if $a$ wasn't $0$, we would have, in (3), a LHS tending to $a \times \infty$ whereas the RHS remains constant, giving a contradiction.

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Second solution involving cartesian equations:

(1) can be expanded in such a way that it appears under the form of a quadratic equation with unknown $y$, considering $x$ as a parameter:

$$(3x)y^2+(-3x^2)y+(x^3-C)=0$$

Its two roots are :

$$y=f(x)=\dfrac{1}{6x} (3 x^2-\sqrt{3x(4C - x^3)})=\dfrac{x}{2}-\sqrt{\dfrac{4C-x^3}{12x}}\tag{3}$$

$$y=g(x)=\dfrac{1}{6x} (3 x^2+\sqrt{3x(4C - x^3)})=\dfrac{x}{2}+\sqrt{\dfrac{4C-x^3}{12x}}\tag{4}$$

The domain of $f$ and $g$ being rather clearly $(0,\sqrt[3]{4C}].$

with blue and green curves resp. below.

Both have a vertical asymptote : the $y$ axis. Indeed, when $x \to 0$, in the second expression of (3), the first fraction tends to $0$, and in the second expression, the fraction under the square root sign is equivalent $\frac{4C}{12x}$ which tends to infinity (Recall: a polynomial behaves like its monomial of lowest degree when $x \to 0$).

(Same reasoning for $g(x)$, of course)