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Homework Help: Hyperbola problem

  1. Dec 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Find an equation of a hyperbola that passes through origin and has asymptotes y = 2x+1 and y= -2x+3


    2. Relevant equations



    3. The attempt at a solution

    I have got the center ( 1/2,2 ) and as it passes through the center i have this equation

    4/a2 - 1/(4b2) from this information how can i get the ratio a/b ?

    Thanks :smile:
     
  2. jcsd
  3. Dec 19, 2009 #2

    Mentallic

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    Draw a diagram first (with reasonable accuracy). Notice that to have the hyperbola pass through the origin with those asymptotes, it must be an "up-down" hyperbola. This means the standard form for the hyperbola that youve used needs to be equal to -1 rather than 1.

    So you have [tex]\frac{4}{a^2}-\frac{1}{4b^2}=-1[/tex]

    And it isn't possible to get the ratio a/b out of this equation. You need to use another piece of information. For a standard hyperbola that has the centre at the origin, what are the equations of its asymptotes?
     
  4. Dec 19, 2009 #3
    I know the fact that it is a up down hyperbola. considering that info and the coordinate of the center i have this form of equation :

    (y-2)2 /a2 - (x-1/2)2/b2 = 1and putting (0,0) in this equation i get this form

    4/a2 - 1/(4b2) = 1

    so i guess it should be right. I have to find now the ratio a/b. And the general form of asymptote lines for a hyperbola in standard position is

    y-k = +- b/a (x-h) where h,k is the coordinate of the center.
     
  5. Dec 19, 2009 #4

    Mentallic

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    Ahh yes sorry for that. Long, embarrassing story short, I mixed up the b and a.

    Ok so what is the ratio of b/a then? If you're unsure, notice the similarity between that formula, and the point-gradient formula for a line:

    [tex]y-k=\pm \frac{b}{a}(x-h)[/tex]

    [tex]y-y_1=m(x-x_1)[/tex]

    Now when you find the ratio of b/a, you can simultaneously solve that with the other equality to find both a and b.


    EDIT:
    I now realize what the little mix up about how the standard form should be set out was. We were both wrong. Again, I'm sorry, I found conics kind of dodgy when I studied them a while back, so now my memory has made it worse. But trust that I'm right now :wink:

    The standard side-side hyperbola with centre at (h,k) is [tex]\left(\frac{x-h}{a}\right)^2-\left(\frac{y-k}{b}\right)^2=1[/tex] This formula is most important and you can adapt the other types from this one!

    But when you're dealing with up-down hyperbolas, it is instead equal to -1.

    So what we have when multiplying through by -1 is [tex]\left(\frac{y-k}{b}\right)^2-\left(\frac{x-h}{a}\right)^2=1[/tex]

    See how the b divides the y coordinate, not the x.
    Now just follow the previous advice I had given, and all will fall into place :smile:
     
    Last edited: Dec 19, 2009
  6. Dec 20, 2009 #5
    so i have to find the gradient of those two lines ? It gives me 2 and -2 for those two lines. Then b/a = +-2 ?
     
  7. Dec 20, 2009 #6

    Mentallic

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    Yes that's right.

    Simultaneously solve

    [tex]\frac{4}{b^2}-\frac{1}{4a^2}=1[/tex]

    and

    [tex]\frac{b}{a}=\pm 2[/tex]

    But you should notice that that when you square both sides it rids you of the [itex]\pm[/itex] so don't be freaked out by it :smile:
     
  8. Dec 20, 2009 #7

    tiny-tim

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    Hi farmd684! Thanks for the PM. :smile:
    Start again …

    since the centre is (1/2,2), the equation must be (x - 1/2)2/a2 - (y - 2)2/b2 = constant,

    so use the ratio a/b from the given asymptotes.
     
  9. Dec 20, 2009 #8
    The focal axis should always be defined as (a) in hyperbola (or not). So in my book all up down hyperbola are defined by y2/a2 - x2 /b2 form. I also know that for a updown hyperbola i have asymptotic ratio defined by a/b whereas it is b/a for a right left. As the coordinates are below both of the asymptotic lines so should't it be a updown hyperbola thus following the pattern y2/a2 ? :confused:

    A similar problem from book which answer is given :

    Asymptotes y=x-2 and y=-x+4 and passes through the origin.
    center ( 1/2,2)
    they have found a/b=2 how they got it ?
     
  10. Dec 20, 2009 #9

    tiny-tim

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    Hi farmd684! :smile:

    I don't know why you're following all these rules …

    a hyperbola (with asymptotes of equal slope) has to be of the form (x - p)2/a2 - (y - q)2/b2 = constant …

    if the constant is positive it'll be up-down, and if it's negative it'll be left-right (or is it the other way round?) …

    just put the numbers in and see what works …

    the constant has to be such that x = y = 0 (in this case) satisfies the equation, and also such that when x = y = ∞, y = ± 2x.

    That's all you need to know! :wink:
     
  11. Dec 20, 2009 #10

    HallsofIvy

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    Look at
    [tex]\frac{x^2}{a^2}- \frac{y^2}{b^}= 1[/tex]
    for very large values of x and y. For "regular size" a and b, both [itex]x^2/a^2[/itex] and [itex]y^2/b^2[/itex] will be very, very large. In fact, that poor little "1" on the right is negligible in comparison.

    That is, for very large x and y, the graph will be very close to the graph of
    [tex]\frac{x^2}{a^2}- \frac{y^2}{b^2}= \left(\frac{x}{a}- \frac{y}{b}\right)\left(\frac{x}{a}+ \frac{y}{b}\right)= 0[/tex]
    which is just the two straight lines corresponding to x/a= y/b and x/a= -y/b or [itex]y= \pm\frac{b}{a}x[/itex].



    As to how your book got b/a= 2 from asymptotes y=x-2 and y=-x+4, I have no idea. Since those lines have slope 1 and -1, it should be b/a= 1, not 2. The two lines intersect, of course, where y= x- 2= -x+ 4 so that 2x= 6, x= 3. Then y= 3-2= -3+ 4= 1. If the asymptotes are y= x-2 and y= -x+ 4, as you say, the center is at (3, 1), not (1/2, 2). Such a hyperbola must be of the form [itex](x-3)^2- (y-1)^2= k[/itex] for some (non-zero) number k. If, in addition to having those asymptotes, the hyperbola passes through the origin, we must have [itex](-3)^2- (-1)^2= 8= k[/itex] so that k is either 10 or -10 giving either [itex](x-3)^2- (y-1)^2= 8[/itex] or [itex](x-3)^2- (y-1)^2= -8[/itex] so that [itex](y-1)^2- (x-3)^2= 8[/itex].

    That is, such a hyperbola must be either
    [tex]\frac{(x-3)^2}{8}- \frac{(y-1)^2}{8}= 1[/tex]
    or
    [tex]\frac{(y-1)^2}{8}- \frac{(x-3)^2}{8}= 1[/tex]
     
  12. Dec 20, 2009 #11
    Thanks for the explanation :smile:
    The book definitely published wrong answer.
    Well you wrote
    so that k is either 10 or -10. It should be 8 or -8 right ?
     
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