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I'm looking to find the asymptotes of a general hyperbola in $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ form, assuming I know the center of the hyperbola $(h, k)$. I came up with a solution, but it's too long for me to be confident that I didn't make a mistake somewhere, so I was wondering if I could run it by someone and see if it works. It's mostly algebraic, and I'm prone to making tiny errors in algebra that throw off the entire problem.

So to start, since we know the center $(h, k)$, we can first translate the hyperbola by $(-h, -k)$ using the transform $x_0 = x - \Delta{x}, y_0 = y - \Delta{y}$ with $\Delta{x} = -h$ and $\Delta{y} = -k$. Assuming $F'$ is the translated $F$, we can divide the entire equation by $-F'$ to put it in the following form:

$$ ax^2 + bxy + cy^2 + dx + ey = 1 $$ With $a = -A'/F'$ and $A'$ the translated $A$, $b = -B'/F'$ and $B'$ the translated $B$, and so on.

Next we convert to polar coordinates to get the following:

$$ r^2(a\cos^2{\theta} + b\cos{\theta}\sin{\theta} + c\sin^2{\theta}) + r(d\cos{\theta} + e\sin{\theta}) - 1 = 0 $$

Solving for $r$ will give us

$$ r = \frac{-d\cos{\theta} - e\sin{\theta} \pm \sqrt{(d\cos{\theta} + e\sin{\theta})^2 + 4(a\cos^2{\theta} + b\cos{\theta}\sin{\theta} + c\sin^2{\theta})}}{2(a\cos^2{\theta} + b\cos{\theta}\sin{\theta} + c\sin^2{\theta})} $$

Now assume $\theta_0 = 2(a\cos^2{\theta} + b\cos{\theta}\sin{\theta} + c\sin^2{\theta})$, this means that as $\theta_0 \rightarrow 0^{\pm}$, $r \rightarrow \pm\infty$. The angles at which $r \rightarrow \pm\infty$ are the asymptotes of the hyperbola, so now it's just a matter of solving for where $\theta_0 = 0$. This is where the majority of the algebra takes place and this is where I'm worried I made some miniscule mistake.

$$ \begin{align} & 2(a\cos^2{\theta} + b\cos{\theta}\sin{\theta} + c\sin^2{\theta}) = 0\\ & \Longleftrightarrow a(1 - \sin^2{\theta}) + b\cos{\theta}\sin{\theta} + c\sin^2{\theta} = 0 \\ & \Longleftrightarrow \sin^2{\theta}(c - a) + b\cos{\theta}\sin{\theta} = -a \\ & \Longleftrightarrow \frac{1 - \cos{2\theta}}{2}(c - a) + \frac{b}{2}\sin{2\theta} = -a \\ & \Longleftrightarrow (c - a)(1 - \cos{2\theta}) + b\sin{2\theta} = -2a \\ & \Longleftrightarrow (c - a)(1 - \cos{2\theta}) + 2a = -b\sqrt{1 - \cos^2{2\theta}} \\ & \Longleftrightarrow \frac{a - c}{b}(1 - \cos{2\theta}) - \frac{2a}{b} = \sqrt{1 - \cos^2{2\theta}} \\ & \Longleftrightarrow (\frac{a - c}{b}(1 - \cos{2\theta}))^2 - 4a\frac{a - c}{b^2}(1 - \cos{2\theta}) + (\frac{2a}{b})^2 = 1 - \cos^2{2\theta} \\ & \Longleftrightarrow (\frac{a - c}{b})^2(1 - 2\cos{2\theta} + \cos^2{2\theta}) - 4a\frac{a - c}{b^2}(1 - \cos{2\theta}) + (\frac{2a}{b})^2 = 1 - \cos^2{2\theta} \\ & \Longleftrightarrow [(\frac{a - c}{b})^2 + 1]\cos^2{2\theta} + [2(\frac{a - c}{b})(\frac{a}{b} - \frac{a - c}{b})]\cos{2\theta} + [(\frac{2a}{b})^2 + (\frac{a - c}{b})^2 - 4a\frac{a - b}{b^2}) - 1] = 0 \\ & \Longleftrightarrow [(\frac{a - c}{b})^2 + 1]\cos^2{2\theta} + 2(\frac{a - c}{b})(\frac{c}{b})\cos{2\theta} + [(\frac{2a}{b})^2 + (\frac{a - c}{b})(\frac{-3a - c}{b}) - 1] = 0 \\ & \Longleftrightarrow [(\frac{a - c}{b})^2 + 1]\cos^2{2\theta} + 2(\frac{a - c}{b})(\frac{c}{b})\cos{2\theta} + [\frac{a^2 + 2ac + c^2}{b^2} - 1] = 0 \\ \end{align} $$ Now let $$ U = [(\frac{a - c}{b})^2 + 1] \\ V = 2(\frac{a - c}{b})(\frac{c}{b}) \\ W = \frac{a^2 + 2ac + c^2}{b^2} - 1 $$ So that the above equation becomes $$ U\cos^2{2\theta} + V\cos{2\theta} + W = 0. $$ Solving for $\cos{2\theta}$ gives us $$ \cos{2\theta} = \frac{-V \pm \sqrt{V^2 - 4UW}}{2U}. $$ Finally we may solve for theta like so, $$ \theta = \frac{1}{2}\arccos{\frac{-V \pm \sqrt{V^2 - 4UW}}{2U}}. $$ This gives us two numbers, $\theta_1$ and $\theta_2$, each corresponding to the slopes $m_1$ and $m_2$ of the asymptotes.

The relationship between the slope of a line $m$ and the angle $\theta$ between the line and the positive x-axis is $m = \tan{\theta}$. You can use the identity $\tan{(\frac{1}{2}\arccos{x})} = \sqrt{\frac{1 - x}{x + 1}}$ to solve for $m_1$ and $m_2$ in terms of non-trig functions, but I think this answer is sufficient enough.

Now that we have the slopes of the asymptotes, we can find the y-intercepts $b_1$ and $b_2$ of each line by simply plugging in the original center $(h, k)$ into each equation for the line and solving.

$$ b_1 = k - m_1h \\ b_2 = k - m_2h $$

Thus, the asymptotes of the hyperbola with general equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ have equations $y = m_1x + b_1$ and $y = m_2x + b2$.

Does this look correct? Also, did I overcomplicate things? Was there an easy solution all along that I was missing?

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  • $\begingroup$ If you’ve translated the origin to the hyperbola’s center, you shouldn’t have any linear terms left in the equation. $\endgroup$ – amd May 13 at 2:06
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So a fairly simple calculus based solution arises from knowing that the Asymptotes are the points where the slope tends toward being constant

That is given

$$Ax^2 + By^2 + Cxy + Dx + Ey + F = 0$$

We begin by deriving with respect to $x$

$$2Ax + 2By \frac{dy}{dx} + Cy+Cx\frac{dy}{dx} + D + E\frac{dy}{dx} = 0 $$

And now solve for $\frac{dy}{dx}$

$$\frac{dy}{dx} = -\frac{2Ax + Cy + D}{2By + Cx + E} $$

Now we can take the limit as $x \rightarrow \pm \infty$ to determine the slopes of the Asymptotes and call this values $m$

Now take the original curve, and solve it for $y$ (it won't be too pretty)

$$y = \frac{-E- Cx \pm \sqrt{(E + Cx)^2 - 4B(Dx + F)}}{2B} $$

And evaluate $$ \lim_{x \rightarrow \pm \infty} y(x) - mx $$

to recover the y intercepts of the asymptotes.

This ENTIRE process can now be repeated with $\frac{dx}{dy}$ in place of $\frac{dy}{dx}$ and y exchanged with x to interpret the asymptotes in a reflected coordinate systems

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    $\begingroup$ OOOH jeez I don't know why I didn't even consider that as a solution it's much less involved than mine :). Though I'm worried about the limit. I'm trying to find a solution which I can code (floating point precision isn't a concern), and I'm not sure how I would program something to evaluate a limit like that. $\endgroup$ – user3002473 Aug 15 '14 at 6:40
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    $\begingroup$ Built in libraries are an option, the other more work intensive option is to basically use if-else to identify if the dy/dx and dx/dy expressions are even defined (think about what happens if A or B or C etc... = 0 to cover all cases) if they are defined its easy to find the limits (L'Hopitals rule by hand will make it pretty obvious what needs to be coded), for the second part finding the limit can just involve increasing $x$ (or y) for expression $y(x) - mx$ to compute a difference $d_i$ and then pick a desired level of accuracy, w, break program and return when $|d_{i+1} - d_i| < w$ $\endgroup$ – frogeyedpeas Aug 15 '14 at 6:49
  • $\begingroup$ When you took the derivative with respect to $x$, you missed applying the product rule to $Cxy$. The correct derivative is $-(2Ax + Cy + D)/(2By + Cx + E)$ $\endgroup$ – Joshua Mundinger Aug 15 '14 at 19:24
  • $\begingroup$ @frogeyedpeas just to clarify, what do you mean when you say $\lim_{x \rightarrow \pm \infty}y(x) - mx$? $y(x) - mx \rightarrow \pm\infty$ for all $y(x)$ and $m$, and no test scenario I've been able to come up with has yielded anything other than $\pm \infty$ as a solution for $b$ (take a look at this example problem: wolframalpha.com/input/…). $\endgroup$ – user3002473 Aug 15 '14 at 20:27
  • $\begingroup$ Hello @frogeyedpeas, i'm in difficult to compute the limit because the coefficients can be positive or negative. Can you do it the step by step please? $\endgroup$ – Gennaro Arguzzi May 12 at 20:08
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For conics $$ax^2+2hxy+by^2+2gx+2fy+\color{blue}{c}=0$$

which is a hyperbola when $ab-h^2<0$.

Its asymptotes can be found by replacing $\color{blue}{c}$ by $\color{red}{c'}$ where

$$\det \begin{pmatrix} a & h & g \\ h & b & f \\ g & f & \color{red}{c'} \end{pmatrix} =0$$

That is

$$\color{red}{c'}=\frac{af^2+bg^2-2fgh}{ab-h^2}$$

The asymptotes are

$$\fbox{$ax^2+2hxy+by^2+2gx+2fy+\frac{af^2+bg^2-2fgh}{ab-h^2}=0 \,$}$$

Alternatively, using the centre of the conics

$$ \left( \frac{bg-fh}{h^2-ab}, \frac{af-gh}{h^2-ab} \right)$$

and the slope $m$ of an asymptote is given by

$$a+2hm+bm^2=0$$

On solving,

$$m=\frac{-h \pm \sqrt{h^2-ab}}{b}=\frac{a}{-h \mp \sqrt{h^2-ab}}$$

Therefore

$$\fbox{$ y-\frac{af-gh}{h^2-ab}= \frac{-h \pm \sqrt{h^2-ab}}{b} \left( x-\frac{bg-fh}{h^2-ab} \right) $}$$

See another answer here for your interest.

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  • $\begingroup$ Hello @Ng Chung Tak I tried to use your equation to compute the asymptotes of the equilateral hyperbola $Ax+By+xy-C=0$, but it does not work, I get $y=\frac{-4Ax+(2B-1)AB}{4(B+x)}$, instead the right ones are $y=-A$ and $x=-B$. Can you help me to understand where I did a mistake please? $\endgroup$ – Gennaro Arguzzi May 13 at 9:16
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    $\begingroup$ @GennaroArguzzi You must have made some computing error: you should get $c'=AB$, which then gives the right equations. $\endgroup$ – Aretino May 13 at 10:52
  • $\begingroup$ @GennaroArguzzi Note that $$\frac{-h \pm \sqrt{h^2-ab}}{b}=\frac{a}{-h \mp \sqrt{h^2-ab}}$$ re-arranging the denominators and putting $a=b=0$ you may get $hx+f=0$ and $hy+g=0$ for each cases. $\endgroup$ – Ng Chung Tak May 13 at 13:34
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    $\begingroup$ @GennaroArguzzi It is $2g=A$. $\endgroup$ – Aretino May 13 at 16:30
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    $\begingroup$ Solve for $x$ instead of $y$. Or, better, factor the equation. $\endgroup$ – Aretino May 13 at 17:10

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