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As given on the Wikipedia page,

an asymptote is a line which becomes the tangent of the curve as the $x$ or $y$ cordinates of the curve tends to infinity.

Hyperbola has asymptotes but parabolas ( both being an open curve and a conic section) do not.

I am just curious to know why don't parabolas have an asymptote ? Is there any mathematical proof to show that ?

Thanks in advance:)

Blue
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Ankit
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  • How do you define a parabola? – Andrew Chin Sep 05 '20 at 05:24
  • @Andrew Chin it is the locus of points whose distance from a fixed line and from a fixed point is equal. – Ankit Sep 05 '20 at 05:40
  • Add a link to the page – Arjun Sep 05 '20 at 05:56
  • Hyperbolas are the only conic sections with asymptotes. Even though parabolas and hyperbolas look very similar, parabolas are formed by the distance from a point and the distance to a line being the same. Therefore, parabolas don’t have asymptotes. – SarGe Sep 05 '20 at 06:58
  • @SarGe this is the line which I couldn't understand. Can you make it clear ? – Ankit Sep 05 '20 at 07:00
  • You might be interested in the last part of this answer to a related question: https://math.stackexchange.com/questions/3621086/is-there-geometric-proof-for-the-equation-of-hyperbola-using-only-constant-dista/3623154#3623154 – Intelligenti pauca Sep 05 '20 at 15:38
  • If we define an asymptote as a tangent to the curve at a point at infinity in the extended Euclidean plane, a parabola has exactly one asymptote. The asymptote is the line at infinity. – Maxim Sep 10 '20 at 18:56

2 Answers2

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If a parabola had an asymptote, then we could translate-rotate it so that the asymptote coincides with the $y$ axis.

Now, consider the general conic section (which, of course, includes all rotated parabolas):

$$A x^2 + B x y + C y^2 + D x + E y + F= 0 \tag1$$ Or $$ y( C y + B x + E) + A x^2 + D x + F=0 \tag2$$

and let's see under what conditions it could happen its graphic has the $y$ axis as an asymptote, i.e. that as $x\to 0$ $y \to \pm\infty$.

In that limit, we must have $$( C y + B x + E) \to -\frac{F}{y} \to 0\tag3$$

But this requires $C=0$ and $E=0$

Then it cannot be a parabola (or an ellipse), only an hyperbola.

leonbloy
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  • Great answer! But for $( C y + B x + E) \to 0$, isn't $y \to \frac{-E}{C}$ sufficient? And for that, the only conclusions I can see are $E \to \pm \infty$, $C \to 0$ or both. How is it only $C=E=0$ that you got as a solution? Admittedly, I've never heard anyone else talk about coefficients being limits, but this is what I can conclude analytically. Where did I go wrong? – harry Aug 26 '21 at 15:40
  • @harry When want both $x\to 0$ and $y \to \infty$ to happen. – leonbloy Aug 26 '21 at 17:20
  • But isn't that the condition that's required? That's what you've used in your answer, right? – harry Aug 26 '21 at 18:17
  • Since arms of a parabola become parallel as distance increases. Shouldn't the two parallel lines that they are approaching/going to become considered the asymptotes of the parabola? – Aurelius May 14 '24 at 18:43
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Hints:

An $\textit{asymptote}$ is a straight line to which the curve approaches while it moves away from the origin. The curve can approach the line from one side, or it can intersect it again and again. Not every curve which goes infinitely far from the origin (infinite branch of the curve) has an asymptote.

The graph of the function $$y=ax^n$$

where $\space n>0, \space$ integer, is a $\textit{parabola of n-th degree, or of n-th order}.$

For functions given in explicit form $\space y=f(x), \space$ we know: the vertical asymptotes are at points of discontinuity where the function $\space f(x)\space$ has an infinite jump; the horizontal and oblique asymptotes have the equation:

$$y=kx+b, \space \text{ with } \space \space k=\lim_{x \to \infty}\frac{f(x)}{x}, \space \space b=\lim_{x \to \infty}\left[f(x)-kx \right].$$

Anton Vrdoljak
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