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Homework Help: Help, stuck Graphing a hyperbola

  1. Feb 15, 2008 #1
    Hi, I am graphing a hyperbola and have completed everything, except I am not to sure how to draw the final curve in. I have the center, foci, vertices, and axis'. I cannot seem to find any information on actually plotting the curve in. Am I supposed to freehand it in from the vertice and follow the the lines drawn of the axis'? I was trying to find x and y coordinates by substituting numbers into the equation, but the center (h,k) is not at zero it is at (2,2) so I do not know how I would get the variables for the other side?

    Here is the equation:

    2 2
    (Y-2) (X-2)
    ______ - _____ = 1
    3 6
    Any help would be appreciated, math is not my strongest subject. Thanks.
  2. jcsd
  3. Feb 15, 2008 #2
    Sorry, in that equation the two 2's are squares and the 3 is supposed to be under the y-2 and the 6 under the x-2.
  4. Feb 15, 2008 #3
    Just plot about 3 points on each side so you get the general curve then just freehand it, until the edge of the graph.
  5. Feb 15, 2008 #4
    Thanks for your help. I will try that.
  6. Feb 16, 2008 #5


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    Generally, you can't be sure where symbols on different lines will be placed on the internet. Either (y-2)^2/3 - (x- 2)^2/6= 1 or use HTML (y-2)2/3- (x-2)2/6= 1 or "LaTex":
    [tex]\frac{(y-2)^2}{3}- \frac{(x-2)^2}{6}= 1[/tex]

    I presume you can see that the center is at (2, 2), the semi-axes are [itex]\sqrt{6}[/itex] and [itex]\sqrt{3}[/itex] and that the vertices are at (2, 2[itex]\pm\sqrt{3}[/itex]).

    Here's how I would graph that. In light pencil, since they aren't part of the graph itself, draw a horizontal line of length [itex]\sqrt{6}[/itex] right and left of (2, 2) and a vertical line of length [itex]\sqrt{3}[/itex] above and below (2,2). Now draw a small rectangle having those as "center lines". That is, two horizontal lines at y= 2+ [itex]\sqrt{3}[/itex] and y= 2- [itex]\sqrt{3}[/itex] and two vertical lines at x= 2+ [itex]\sqrt{6}[/itex] and x= 2- [itex]\sqrt{6}[/itex]. Finally, draw the diagonals of that rectangle, extending the as far as you can. Those are the "asymptotes" of the hyperbola.

    Now, you are ready to draw the hyperbola! You can see from the original equation, [itex](y-2)^2/6-(x-2)^2/6= 1[/itex], that if x= 0, you can solve for the vertices y= 2[itex]\pm\sqrt{3}[/itex] but if y= 0, you cannot solve for x (you get [itex](x-2)^2= -6[/itex]. That tells you that the hyperbola is in the top and bottom wedges formed by the asymptotes.

    Start at the upper vertex and draw smooth curve out on each side to the asymptotes, getting closer and closer to them. Do the same at the lower vertex.
  7. Feb 16, 2008 #6
    Thank you, Hallsofivy. I have followed what you explained so clearly. The thing that I don't understand is that math is such an exact science that I thought there must be a special formula for determining exact points to follow to draw the hyperbola. But from the suggestions given here it appears that I am over analyzing and that all I need to do is free hand it. Thanks a lot for your help.
  8. Feb 16, 2008 #7


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    There is! It's the formula for the hyperbola!

    Or perhaps you mean a formula that say y= ... for any value of x.

    Since your equation is (y-2)2/3 - (x-2)2/6= 1, (y-2)2/3= 1+ (x-2)2/6 so (y-2)2= 3+ (x-2)2/2 and then
    [itex]y= 2\pm\sqrt{3+ (x-2)^2/2}[/itex]
    Is that better? On a graphing calculator, for example you could enter that with the "+" sign in one function, with the "-" sign in another and graph the entire hyperbola.
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