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It's known that the asymptotes of a hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is given by $y=\pm\frac{b}{a}x$ if $a>b$.

I tried to find a proof of the fact that why the equations of these asymptotes are like that,however the only reference (Thomas calculus book) that I found explained that the two asymptotes are derived by letting $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=0$.


It would be highly appreciated if someone prove why the equation of the asymptotes have such form.

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    $\begingroup$ Does this answer your question? Asymptotes of hyperbola $\endgroup$ – devianceee May 30 at 10:20
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    $\begingroup$ If your attitude is to blame your books, I fear for your mathematical future. $\endgroup$ – TonyK May 30 at 11:37
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    $\begingroup$ I think we can improve this question by removing the third paragraph. I'm not sure if it serves any purpose other than as a rant, which is inappropriate for SE. $\endgroup$ – Bladewood May 30 at 18:33
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    $\begingroup$ @ Bladewood, If saying the truth is considered as inappropriate , I will stop contributing to SE,Of course some attitudes are defined,but "It is not people who break ethical standards who are regarded as aliens. It is people like me who are isolated." Grigori Perelman. $\endgroup$ – user794034 May 30 at 19:34
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    $\begingroup$ number, you are no Grigori Perelman. $\endgroup$ – TonyK May 30 at 19:49
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Consider the focus-directrix definition of the hyperbola:

Given a (focus) point $F=(c,0)$ and a (directrix) line $\delta: x=d$, say, with $0<d<c$, the hyperbola is the locus of points $P$ such that $$\text{distance from $F$}=\text{eccentricity}\cdot(\text{distance from $\delta$})$$ for some $\text{eccentricity}$ $e > 1$.

For $P$ really-really-really-really-$\cdots$-really far away from the origin, its distance to $F$ is virtually-indistinguishable from its distance to the origin; and its distance from $\delta$ is virtually-indistinguishable from its distance to the $y$-axis. This makes $P$ virtually-indistinguishable from a point $Q$ travelling on a locus defined by $$\text{distance from $O$} = \text{eccentricity}\cdot(\text{distance from $y$-axis})$$ The equation for $Q$'s locus is $$\sqrt{x^2+y^2}=e x=\frac{c}{a}x=\frac{\sqrt{a^2+b^2}}{a}x \quad\to\quad x^2+y^2=\frac{a^2+b^2}{a^2}x^2\quad\to\quad \frac{x^2}{a^2}-\frac{y^2}{b^2}=0$$ (with $a:=c/e$ and $b:=\sqrt{c^2-a^2}$, which (one can show) match our common interpretations of these values), and we recognize this as representing a pair of crossed lines. We see, then, that on a grand scale, the hyperbola approaches these lines, which we accordingly call its asymptotes. $\square$

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Edited to do it properly -- see below

Original post:

We have $$y=b\sqrt{\frac{x^2}{a^2}-1}=\frac{b}{a}x\sqrt{1-\frac{a^2}{x^2}}$$ And as $x\to\pm\infty$, $\sqrt{1-\frac{a^2}{x^2}}\to 1$.

End of original post

But as mentioned in the comments, it is not enough to show that $\frac{y}{bx/a}\to 1$. We have to show that $y-\frac{b}{a} x\to 0$:

$$y-\frac{b}{a}x=\frac{b}{a}x\left(\sqrt{1-\frac{a^2}{x^2}}-1\right)$$ But $$1-\frac{a^2}{x^2}\le\sqrt{1-\frac{a^2}{x^2}}<1$$ So $$\left|\sqrt{1-\frac{a^2}{x^2}}-1\right|\le\frac{a^2}{x^2}$$ Therefore $$\left|y-\frac{b}{a}x\right|\le\frac{b}{a}|x|\cdot\frac{a^2}{x^2}=\frac{ba}{|x|}$$ which tends to $0$ as $x\to\pm\infty$.

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    $\begingroup$ Thanks fo the answer,you say $\lim_{x \to +\infty} \sqrt{1-\frac{a^2}{x^2}}=1$,But why we can't say $$\lim_{x \to +\infty}y=\lim_{x \to +\infty}\frac{b}{a}x\sqrt{1-\frac{a^2}{x^2}}=\lim_{x \to +\infty}\frac{b}{a}x\lim_{x \to +\infty}\sqrt{1-\frac{a^2}{x^2}}=+\infty$$? $\endgroup$ – user794034 May 30 at 13:34
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    $\begingroup$ @number: Since $y$ does get infinitely-large along with $x$, the infinite limiting statement is true. The question is, though: How does $y$ compare to $x$ in the long run? So, perhaps a better way to capture this is $$\lim_{x\to\infty}\frac{y}{x}=\lim_{x\to\infty}\frac{b}{a}\sqrt{1-\frac{a^2}{x^2}} = \frac{b}{a}\cdot \lim_{x\to\infty}\sqrt{1-\frac{a^2}{x^2}}=\frac{b}{a}\cdot 1$$ $\endgroup$ – Blue May 30 at 20:52
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    $\begingroup$ Actually, @Blue, we are both wrong: it is not enough that $\dfrac{y}{(b/a)x}$ tends to $1$; for an asymptote, we need that $y - (b/a)x$ tends to $0$. $\endgroup$ – TonyK May 30 at 22:38
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    $\begingroup$ @TonyK Why is that? Do you have an example? $\endgroup$ – gen-z ready to perish May 30 at 22:50
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    $\begingroup$ @gen-zreadytoperish: you confused me there. I though you were responding to my response to your comment, but no. Anyway: For a function to have an asymptote, it must approach that asymptote arbitrarily closely in absolute terms, not just in relative terms. So although my post shows that $\frac{b}{a}x$ is the only asymptote possible, it doesn't show that $\frac{b}{a}x$ is necessarily an asymptote -- it might be that there is no asymptote. For instance, $f(x)=x+\sin x$. $\endgroup$ – TonyK May 30 at 23:09
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The other answers have tried to give more rigorous arguments, which I would like to complement with a heuristic way which doesn't need further manipulation of the equation:

From looking at the hyperbola, it is obvious that the asymptotes are lines that the curve approaches when $x$ and $y$ become very large, in particular larger than $a$ or $b$ (BTW, the condition $a<b$ seems unnecessary to me). Then, in the defining equation$$\frac{x²}{a²}-\frac{y²}{b²}=1\,,$$ you have two large numbers on the left-hand side whose difference is $1$. In other words, their difference is much smaller than the numbers themselves, and it becomes a good approximation to just neglect the $1$ on the right-hand side. Furthermore, the approximation becomes better which increasing $x$ and $y$. Thus, $$\frac{x²}{a²}-\frac{y²}{b²}=0\,,$$ is at least a good candidate for the equation of the asymptotes.

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  • $\begingroup$ This is a great heuristic, the type of which I’ve seen many times in physics. We intelligently (usually with foresight) replace some expressions with simpler ones, so the equations can be analytically solved. $\endgroup$ – Bob Krueger May 30 at 19:47
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Consider a family of hyperbolas

$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=t,$$ for any real number $t\ne 0.$ When $t=0,$ this just gives a pair of intersecting straight lines.

By a linear transformation (which does not change the character of the conic), you can consider the affine family of hyperbolas $$xy=t$$ instead. Then considering $y$ as a function of $x$ gives $$y=\frac tx,$$ where we may take $t>0$ without loss of generality.

We know that as $x\to\pm \infty,$ then $y\to 0.$ (By a similar reasoning, $x=0$ when $y=\pm\infty.$)

Therefore, it follows that the equation $y=0$ is asymptotic to $y=t/x,$ and that $x=0$ is asymptotic to $x=t/y.$

In general, each member of the family is asymptotic to the pair of lines obtained when $t=0.$ (The so-called degenerate case.)

Hence, our original family $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=t,$$ when we invert the transformation, must also be asymptotic to the lines obtained when we set $t=0.$ This gives the result.

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Aymptotes of a hyperbola are a limiting case of tangents which tend to meet the hyperbola at $\infty$.So take a general equation of line $y=mx+c$ and plug it into the hyperbola equation.This gives you $$\frac{x^2}{a^2}-\frac{(mx+c)^2}{b^2}=1$$ which gives $$x^2(\frac{1}{a^2}-\frac{m^2}{b^2})-\frac{2mcx}{b}-\frac{c^2}{b^2}-1=0$$ and we know that this eqaution should have both roots tending to infinity(since i said aymptotes of a hyperbola are just tangents which meet the hyperbola at $\infty$).Now applying conditions for roots tending to $\infty$ for a quadratic equations gives coeffecient of $x^2=0,x=0$ and $constant\neq 0$(this can be proved by assuming $\alpha$, $\beta$ as roots of equation $px^2+qx+r$ and both should tend to zero,which means $\frac{1}{\alpha},\frac{1}{\beta}$ should tend to zero,now forming a quadratic equation with $\frac{1}{\alpha},\frac{1}{\beta}$ gives us $rx^2+qx+p=0$ which has both roots tending to zero if $p\to 0,q\to 0$ and $r\neq o$) and thus we get $$\frac{1}{a^2}-\frac{m^2}{b^2}=0\space and\space \frac{2mc}{b}=0$$ which gives us $$m=\pm\frac{b}{a} and\space c=0$$ and which on substituting in $y=mx+c$ and multiplying both equations gives the famous pair of asymptotes formula as you said $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=0$$ Just as a side note; this method can be applied to find the asymptotes of any curve (even twisted, translated and rotated hyperbolae for that matter). Hope that helps!

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    $\begingroup$ Thanks you for the answer,can you explain why do we need to plug $y=mx+c$ into the equation of the hyperbola? $\endgroup$ – user794034 May 30 at 13:36
  • $\begingroup$ We are taking trying to find a pair of lines which touch the graph as we move further away from its centre, (AKA asymptotes). As these lines are no different from any of the other lines in the plane in terms of algebra, ie every line can be written as y = mx +c, we can assume the line to be y = mx +c and for particular values for m and c, this line will be our asymptote. We find those particular values by applying conditions for the roots.It's just like finding intersection points of two equations by substituting one in other(in algebraic perspective) $\endgroup$ – Thenard Rinmann May 30 at 13:43

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