# finding hyperbola asymptotes

Given implicit function F(x,y) = 0, how can I find its asymptotes?

EDIT: Sorry, my calculations were wrong. Here is correct function:

$F(x,y)=\sqrt{(x-a)^2 + (y-b)^2} - \sqrt{(x-c)^2 + (y-d)^2} - e$

-
@J.M. a, b and c are real numbers. –  qutron Dec 27 '10 at 14:26
@J.M. 20000 - 100000 –  qutron Dec 27 '10 at 14:43
The zero set of the function $F$ you wrote in your question is not a hyperbola: the value $F(x,y)$ depends only on $\sqrt{x^2+y^2}$, so the zero set has rotational symmetry around the origin. –  Mariano Suárez-Alvarez Dec 27 '10 at 15:18
@Mariano Suárez-Alvarez. I might be wrong. I will check my calculations. –  qutron Dec 27 '10 at 15:35
So, only four parameters? –  Ｊ. Ｍ. Dec 27 '10 at 15:46

After rationalization, your implicit equation can be expressed in standard quadratic form as

$$px^2+qxy+ry^2+sx+ty+u=0$$

where

$$p=4(a-c)^2-4e^2$$
$$q=8(a-c)(b-d)$$
$$r=4(b-d)^2-4e^2$$
$$s=4(a+c)e^2-4(a-c)\left(a^2+b^2-c^2-d^2\right)$$
$$t=4(b+d)e^2-4(b-d)\left(a^2+b^2-c^2-d^2\right)$$
$$u=e^4-2\left(a^2+b^2+c^2+d^2\right)e^2+\left(a^2+b^2-c^2-d^2\right)^2$$

The earlier version of my answer would now apply, but since Isaac has written a clearer way of doing it, I'll demonstrate the alternative "dumb" approach ("dumb" in the sense of not exploiting the structure of the problem).

Again, before you go through the trouble of doing these computations, you'd first want to ensure that the discriminant $\Delta=64e^2((a-c)^2+(b-d)^2-e^2)$ is negative.

Having done so, the idea is as follows: replace all instances of $y$ in the implicit equation $F(x,y)=0$ with $mx$, divide by $x^2$ (since the degree of the algebraic equation is 2), and let $x\to\infty$. This yields the quadratic equation

$$rm^2+qm+p=0$$

which should be solved for $m$; the initial check of the discriminant would ensure that both roots of that quadratic in $m$ are real.

Letting $m_i$ be any of the two roots $m_1$ and $m_2$ of that quadratic, one now goes back to the original implicit equation, and replaces $y$ with $m_i x+k_i$. From this, note that the quadratic terms disappear, and only the linear terms remain. Divide by $x$, let $x\to\infty$, equate the remaining terms to 0, and solve for the appropriate value of $k_i$. If you do that, the expression you should get for $k_i$ is

$$k_i=-\frac{tm_i+s}{2rm_i+q}$$

from which $y=m_1 x+k_1$ and $y=m_2 x+k_2$ are your two asymptotes.

-
Thank You. I got the idea, but F(x) is a little bit different. –  qutron Dec 27 '10 at 13:43
I might be wrong. I will check my calculations. Anyway, thanks)) –  qutron Dec 27 '10 at 15:34
Thank You! I appreciate it. –  qutron Dec 28 '10 at 14:23

The points $(a,b)$ and $(c,d)$ are the foci of the hyperbola and the hyperbola is the locus of points for which the difference of the distances to the foci is $|e|$, though because your $F(x,y)=0$ is only looking at the difference in one direction, your graph is probably only half of the hyperbola (the other half being $0=\sqrt{(x-a)^2+(y-b)^2}-\sqrt{(x-c)^2+(y-d)^2}+e$).

The asymptotes of the hyperbola pass through the center, which is the midpoint of the segment joining the foci, $$\left(\frac{a+c}{2},\frac{b+d}{2}\right).$$

Let $f=\sqrt{(a-c)^2+(b-d)^2}$, the distance between the foci, and let $g=\sqrt{f^2-e^2}$. If the foci had the same $x$-coordinate (that is, if the hyperbola were horizontally-oriented), the slopes of the asymptotes would be $\pm\frac{g}{e}$, so the acute angles the asymptotes make with the line through the foci are $\pm\arctan\frac{g}{e}$. The acute angle the line through the foci makes with the $x$-axis is $\arctan\frac{b-d}{a-c}$, so the angles from the $x$-axis to the asymptotes are $\arctan\frac{b-d}{a-c}\pm\arctan\frac{g}{e}$, and the slopes of the asymptotes are $$\tan\left(\arctan\frac{b-d}{a-c}\pm\arctan\frac{g}{e}\right)=\frac{be-de+ag-cg}{ae-ce-bg+dg}\text{ or }\frac{be-de-ag+cg}{ae-ce+bg-dg}.$$

With a point and the slopes, you should be able to write equations of the asymptotes.

-
Thanks a lot! –  qutron Dec 28 '10 at 14:24