Asymptote of $x^2 + y^2 -3xy = 1$

Question

How can I find the asymptote of the hyperbola $x^2 + y^2 -3xy = 1$?

I tried to convert the equation to $y=±\sqrt{1+3xy−x^2}$ and mark $x=\infty$ and I got $y=\pm\sqrt{3xy−x^3}$ but I don't know how to continue.

Answer 1

Hint:

• assume that $x$ goes to infinity while $y$ remains finite (and symmetrically);

• assume that both $x$ and $y$ go to infinity.

In all these cases, simplify the equation by discarding the terms that become negligible and solve.

Answer 2

We obtain $$y=\frac{3x+\sqrt{(9x)^2-4(x^2-1)}}{2}$$ or $$y=\frac{3x-\sqrt{5x^2+4}}{2}.$$ Now, we'll work with $y=\frac{3x+\sqrt{5x^2+4}}{2}.$

For $y=\frac{3x-\sqrt{5x^2+4}}{2}$ we'll get a same result.

$$\lim_{x\rightarrow+\infty}\frac{y}{x}=\frac{1}{2}\lim_{x\rightarrow+\infty}\left(3+\sqrt{5+\frac{4}{x^2}}\right)=\frac{3+\sqrt5}{2}$$ and $$\lim_{x\rightarrow+\infty}\left(y-\frac{3+\sqrt5}{2}x\right)=\lim_{x\rightarrow+\infty}\frac{\sqrt{5x^2+4}-\sqrt5x}{2}=\lim_{x\rightarrow+\infty}\frac{2}{\sqrt{5x^2+4}+\sqrt5x}=0.$$ Id est, $y=\frac{3+\sqrt5}{2}x$ is an asymptote.

Now, $$\lim_{x\rightarrow-\infty}\frac{y}{x}=\frac{1}{2}\lim_{x\rightarrow-\infty}\left(3-\sqrt{5+\frac{4}{x^2}}\right)=\frac{3-\sqrt5}{2}$$ and $$\lim_{x\rightarrow-\infty}\left(y-\frac{3-\sqrt5}{2}x\right)=\lim_{x\rightarrow-\infty}\frac{\sqrt{5x^2+4}+\sqrt5x}{2}=\lim_{x\rightarrow-\infty}\frac{2}{\sqrt{5x^2+4}-\sqrt5x}=0.$$ Id est, $y=\frac{3-\sqrt5}{2}x$ is an asymptote.

Answer 3

Instead of looking at it as the graph of the functions, you can see it as a conic curve. The asymptotes can be found by applying an enlarging transformation of infinitesimal factor. Let $(X,Y)=(kx,ky)$ which means the enlargement of factor k. Multiplying both sides of the original equation by $k^2$, you get to $X^2+Y^2-3XY=k^2$. Then you can set the factor infinitesimal, i.e. $k\rightarrow0$ to have $X^2+Y^2-3XY=0$. This equation represents a pair of asymptotes. By factorising LHS as $(Y-(3+\sqrt5)X/2)(Y-(3-\sqrt5)X/2)=0$ you have a set of two equations $Y=(3\pm\sqrt5)X/2$.

Answer 4

Hint If the equation of hyperbola is $(x/a)^2 -(y/b)^2=1$, then the combined equation of the asymptotes is $(x/a)^2 -(y/b)^2=0$