Why is the hyperbola round at the bottom?

  1. Jul 14, 2014 #1
    My question is somewhat dumb but for some reason I haven't been able to come up with an answer.

    Why is the hyperbola round at the bottom? Namely, I'm thinking of any of these two equations x^2-y^2=1 or y^2-x^2=1

    These two behave like a line as you approach infinity but then becomes round at the bottom. Why do they behave roundy at the bottom?

    edit: I completely understand why this must be the case if I consider the hyperbola as a conic section. But I'm specifically talking about the equation for a hyperbola.
    Last edited: Jul 14, 2014
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  3. Jul 14, 2014 #2


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    What kind of answer would you expect to this why question?

    Why questions are difficult to answer, because I don't know what answer you want. Is just graphing it with a calculator enough reason for you? Presumably not? Do you want a formal proof of roundness? What does roundness mean anyway?

  4. Jul 14, 2014 #3


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    If we consider ##y^2 = x^2 + 1##, and constrain ##y > 0## (the upper branch of the hypberbola), then this is equivalent to
    $$y = \sqrt{x^2 + 1}$$
    or equivalently
    y-1 &= \sqrt{x^2 + 1} - 1\\
    &= (\sqrt{x^2 + 1} - 1)\left(\frac{\sqrt{x^2 + 1} + 1}{\sqrt{x^2 + 1} + 1}\right) \\
    &= \frac{x^2}{\sqrt{x^2 + 1} + 1}\\
    For small ##x##, the denominator is approximately ##2##, so we get
    $$y - 1 \approx \frac{x^2}{2}$$
    $$y \approx \frac{x^2}{2} + 1$$
    In other words, for small ##x##, the shape is approximately parabolic.
  5. Jul 14, 2014 #4


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    For small x you could simply begin to expand [itex] y = \sqrt{\, x^2 + 1} [/itex] with the general binomial theorem and get the same result.
    Last edited: Jul 14, 2014
  6. Jul 14, 2014 #5


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    Or equivalently, compute the Taylor series. I wasn't sure how much background the OP had, though.

    By the way, just to add to my previous post, if we're uneasy about how much error we introduce by approximating the denominator as ##2##, this can be quantified more precisely. For example, for ##-1/2 \leq x \leq 1/2## we have ##0 \leq x^2 \leq 1/4## and so
    $$2 \leq \sqrt{x^2 + 1} + 1 \leq \sqrt{5/4} + 1 \approx 2.118$$
    Therefore for ##-1/2 \leq x \leq 1/2##, we can bound (the upper branch of) the hyperbola above and below by two parabolas:
    $$\frac{x^2}{2.118} \leq y-1 \leq \frac{x^2}{2}$$
    or equivalently,
    $$\frac{x^2}{2.118} + 1 \leq y \leq \frac{x^2}{2} + 1$$

    If the OP knows some calculus, here are a couple of concepts which may be of interest:

    curvature - roughly speaking, measures "roundness" at each point on a curve by identifying a circle which most closely matches the curve at that point. Smaller radius = higher curvature, and vice versa. A circle has constant curvature at every point, so one way to express the notion that a curve is "round" in some region would be to compute the curvature in that region and find that it is relatively constant (and nonzero).

    smoothness - another measure of how well behaved the curve is: "smooth" = free of "corners" or other sudden behaviors
    Last edited: Jul 14, 2014
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