Asymptotes of the hyperbola $\frac{1}{2} + \frac{y}{5} + \lambda x(\frac{1}{2} + \frac{x}{10} + \lambda y) = 0$.

Question

I am trying to find the asymptotes of the hyperbola $\frac{1}{2} + \frac{y}{5} + \lambda x(\frac{1}{2} + \frac{x}{10} + \lambda y) = 0$.

I read through the answer here , but the equation of the hyperbola involves $x,x^2,y,y^2$ and constants but the equation I have written has a term of $xy$ too, due to which I am unable to convert the equation into the standard form.

Any help on how to find the asymptotes of the hyperbola.

Answer 1

One can express $y$ as a function of $x$ (I have taken $a$ instead of $\lambda$) :

$$y =-\dfrac{ax^2 + 5ax + 5}{10 a^2 x + 2}\tag{1}$$

Thus you have a vertical asymptote with equation $x=-\tfrac{1}{5a^2}$.

The equation of the second asymptote is obtained thanks to the decomposition of (1) into partial fractions :

$$y=Ax+B+\dfrac{C}{10 a^2 x + 2}$$

$y=Ax+B$ will be the equation of this slant asymptote because $\dfrac{C}{10 a^2 x + 2}$ tends to $0$ when $|x| \to \infty$.

One finds $A=-\frac{1}{10a}$ and $B=-\frac{1}{2a}+\frac{1}{50a^3}$.

Fig. 1 : Cases $a=-3$ to $a=3$ with a step $0.5$. The only case where one hasn't a hyperbola is $a=0$ : it is the horizontal line $y=\tfrac{5}{2}$.

Remark : you couldn't get a formula of the form :

$$\dfrac{(x-x_0)^2}{a^2}-\dfrac{(y-y_0)^2}{b^2}=1$$

This form is valid, up to a rotation and a translation, iff asymptotes make a 90° angle, which is not the case here.

Answer 2

An "algorithmic" approach:

Write $y$ as: $$y = -\frac{\lambda x^2 + 5 \lambda x + 5}{10 \lambda^2 x + 2}$$

1. Observe that $$\lim_{x \to +\infty} y$$ and $$\lim_{x \to -\infty} y$$ are not finite. Hence no horizontal asymptotes.

2. Consider the root(s) of the denominator $$10 \lambda^2 x + 2 = 0 => x = -\frac{1}{5\lambda^2}$$ Hence only one vertical asymptote.

3. If any oblique asymptotes exist, they must be of the following form: $$y = mx + c$$ where $$m = \lim_{x \to \infty} \frac{y}{x}$$ and $$c = \lim_{x \to \infty} (y - mx)$$

In this case, there is only one oblique asymptote $$y = mx + c$$

If you calculate the limits, you will find $$m = -\frac{1}{10 \lambda}$$ and $$c = \frac{1-25 \lambda^2}{50 \lambda^3}$$

Answer 3

Working with homogeneous coordinates in the extended Euclidean plane, the asymptotes of the hyperbola are lines that pass through its center and its intersections with the line at infinity. The latter can be found by homogenizing the equation of the hyperbola into $$\frac\lambda{10}x^2+\frac12w^2+\lambda^2xy+\frac\lambda2xw+\frac15yw=0$$ and then setting $w=0$. The two solutions of the resulting equation are $x=0$ and $10\lambda y=-x$, from which the direction vectors of the two asymptotes are $(0,1)^T$ and $(10\lambda,-1)$. The center of the hyperbola can be found in various ways such as setting the partial derivatives of the left-hand side of the equation to zero, giving the point $\left(-{1\over5\lambda^2},{2-25\lambda^2\over50\lambda^3}\right)$. The asymptotes therefore have the equations $$x = -\frac1{5\lambda^2} \\ x+10\lambda y = \frac1{5\lambda^2}-5.$$

Equivalently, one can write the hyperbola’s equation in matrix form as $(x,y,1)Q(x,y,1)^T=0$, with $$Q=\begin{bmatrix}\frac\lambda{10}&\frac{\lambda^2}2&\frac\lambda4\\\frac{\lambda^2}2&0&\frac1{10}\\\frac\lambda4&\frac1{10}&\frac12\end{bmatrix}.$$ The homogeneous coordinates of its center are given by the last row/column of $Q^{-1}$ (the center is the pole of the line at infinity) and its intersections with the line at infinity can be found by solving $(x,y,0)Q(x,y,0)^T=0$ for $x$ and $y$.