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I have so many questions about the Hyperbola

  1. Jun 19, 2011 #1
    1. The problem statement, all variables and given/known data

    I`ll try to make this as orderly as possible, but I've got so many questions about it

    1. The most "general" form of a hyperbola are

    [tex]\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1[/tex]

    [tex] \frac{y^2}{b^2}- \frac{x^2}{a^2}= 1[/tex]

    Now my question is, the first one opens with the x-axis, the second one opens with the y-axis. My question is, I am never going to be able to rememeber them, even if i draw out my asympotetes I am not going ot be able to deduct with which axis does the hyperbola open.

    Also just another side question, the asympotetes are negatives of each other, but when I graphed it, they are also perpendicular to each other. Now here is the thing, how come they aren't negative reciprocal of each other?

    2. Sometimes we call [tex]xy = 1[/tex] as a hyperbola how do I convert those from (1) to this form?

    3. How do the hyperbolic functions apply to (2)?

    4. I never understood this, that's say [tex]\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1[/tex]

    Now I am going to rewrite it as [tex]y = \pm \frac{b}{a}\sqrt{x^2 - a^2}[/tex]

    Now I just want to look at [tex]\sqrt{x^2 - a^2}[/tex]

    How do I recognize that [tex]\sqrt{x^2 - a^2}[/tex] will give me a curve and not a straight line? I used to think that the square and the square root "cancels" and the a doesnt' matter. I was wrong.
  2. jcsd
  3. Jun 19, 2011 #2


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    These formulas are not the most general forms for hyperbolas. These equations correspond to hyperbolas that either open horizontally and are symmetric about the x-axis, or open vertically and are symmetric about the y-axis. You can obviously have hyperbolas which are not symmetric about either axis.

    To figure which equation corresponds to the vertical or horizontal hyperbola, set x=0 and see if there's a real solution to the equation. If there is, that means the graph crosses the y-axis, which means... I'll let you think about that.
    I don't know what this means.
    You can't. The graph of xy=1 is not symmetric about the x-axis or y-axis, so you can't write them in the forms you mentioned in (1).

    The general form for a conic section is
    where at least one of A, B, and C is not 0. If [itex]B^2-4AC>0[/itex], you get a hyperbola. (Other cases correspond to the other conic sections, the ellipse and the parabola.)

    If A=±1/a2, C=∓1/b2, B=D=E=0, and F=-1, you get the forms in (1). For xy=1, you have A=C=D=E=0, B=1, and F=-1.
  4. Jun 19, 2011 #3


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    Actually, the 2nd equation should be
    [tex] \frac{y^2}{a^2}- \frac{x^2}{b^2}= 1[/tex]

    And anyway, as vela said, these aren't the "general" forms. Maybe I would call these the "basic" forms, because the centers are at (0, 0).

    Think of it this way:
    - if the a2 term is underneath the x2 term, then the focal axis (a line that contains the center, vertices, and foci) is horizontal, which means that the two curves open left and right.

    - if the a2 term is underneath the y2 term, then the focal axis is vertical, which means that the two curves open up and down.

    Pretty easy to remember.

    It may have been a coincidence that they were perpendicular, but normally they aren't. The slopes of the asymptotes are
    [tex]\pm \frac{b}{a}[/tex] if the focal axis is horizontal, or
    [tex]\pm \frac{a}{b}[/tex] if the focal axis is vertical.
    The only way that the two asymptotes are perpendicular is if a = b.

    Uh, it's the minus a2 in the square root that prevents
    [tex]\sqrt{x^2 - a^2}[/tex]
    from being a straight line. You can't just cancel out a square with a square root if there are additional terms in the square root. You should also know that the graph of
    [tex]y = \sqrt{x^2 - a^2}[/tex]
    , where a is a constant, is a semicircle. But why would want to look at this alone anyways?

    Try looking at this link below and see if it helps.
    http://www.purplemath.com/modules/hyperbola.htm" [Broken]
    Last edited by a moderator: May 5, 2017
  5. Jun 19, 2011 #4


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    You're thinking of
    [tex]y = \sqrt{a^2 - x^2}[/tex]
    This is the upper half of a circle of radius a, centered at (0, 0).

    If you square both sides of the following equation -
    [tex]y = \sqrt{x^2 - a^2}[/tex]

    you get y2 = x2 - a2, or equivalently (skipping a step or two),
    x2 - y2 = a2
    This is a hyperbola that opens to the left and right, with center at (0, 0), and foci vertices at (+/-a, 0).
  6. Jun 19, 2011 #5


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    Yes, of course. I need to stop posting late at night, damnit. :cry:
  7. Jun 20, 2011 #6


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    Let [itex]x= (\sqrt{2}/2)x'+ (\sqrt{2}/2})y'[/itex] and [itex]y= (\sqrt{2}/2}x'- (\sqrt{2}/2)y'[/itex]

    [tex]xy= ((\sqrt{2}/2)x'+ (\sqrt{2}/2})y')((\sqrt{2}/2}x'- (\sqrt{2}/2)y')= \frac{x'^2}{2}- \frac{y'^2}{2}= 1[/tex]
    That's a hyperbola with x'= y' and x'= -y' as asymptotes so, in the xy-plane, a hyperbola with the x and y axes as asymptotes.

  8. Jun 21, 2011 #7
    Does it make sense that the hyperbola is actually made of two parabolas? It just kinda looks like it...
  9. Jun 21, 2011 #8


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    No, definitely not. The hyperbola is asymptotic to some straight lines but the parabola doesn't have this property.
  10. Jun 22, 2011 #9
    Does a Hyperbola (eventually) behave like a line?
  11. Jun 22, 2011 #10


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    In a sense. It is asymptotic to a line just in the same way y=1/x is asymptotic to the axes.
  12. Jun 22, 2011 #11
    Let's say I have

    [tex]\frac{(x-2)^2}{25} - \frac{(y-1)^2}{9} = 1[/tex]


    [tex] y = \pm \frac{3}{5} \sqrt{(x-2)^2 -5^2} + 1[/tex]

    Now here is the thing, as [tex]x\to \pm \infty[/tex], shouldn't the 2 in (x-2) disappear? But it doesn't why?? Same with the +1
  13. Jun 22, 2011 #12


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    I have no idea what you mean by that. As x goes to infinity, so does x- 2.

    In order to take that limit, the simplest thing to do is to multiply and divide the equation by first term by x- 2, taking the denominator into the root:
    [tex]y= \pm\frac{3}{5}(x- 2)\sqrt{1- \left(\frac{5}{x-2}\right)^2}+ 1[/tex]

    Of course, y still goes to infinity as x does but now we can say that for very large x, the [itex](5/(x-2))^2[/itex] term will be negligible and we we have, approximately,
    [tex]y= \pm\frac{3}{5}(x- 2)+ 1= \frac{3}{5}x- \frac{1}{5}[/tex]
    using the "+" or
    [tex]y= -\frac{3}{5}x+ \frac{11}{5}[/tex]
    using the "-".

    In my opinion, the simplest way to find asymptotes is to start from the original form,
    [tex]\frac{(x- 2)^2}{25}- \frac{(y- 1)^2}{9}= 1[/tex]
    and argue that, for very large x and y, the two fractions, separately are so large compared with "1" that the right side can be ignored:
    [tex]\frac{(x-2)^2}{25}- \frac{(y-1)^2}{9}= 0[/tex]

    [tex]\frac{(x- 2)^2}{25}= \frac{(y-1)^2}{9}[/tex]

    [tex]\pm\frac{x-2}{5}= \frac{y-1}{3}[/tex]

    [tex]\pm 3(x- 2)= 5(y- 1)[/tex]

    which gives the same formula as before.
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