Where does the equation of asymptotes of a hyperbola come from?

Question

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 $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=0$.

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

Answer 1

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 $x\to\pm\infty$.

Answer 2

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.

Answer 3

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!

Answer 4

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.

Answer 5

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. $\square$