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Homework Help: Find the equation of the hyperbola with centre at the origin

  1. Mar 7, 2005 #1
    Find the equation of the hyperbola with centre at the origin and sketch the graph.
    e. tranverse axis is on the y-axis and passes through the points R(4, 6) and S(1, -3)

    How would I find a and b? I plugged in the coordinates in [tex]\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1[/tex] and came up with two equations. The thing is, I get to the point where I have to take the square root of both sides, but one side is always negative. I thought you couldn't take the square root of a negative? :confused:
     
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  3. Mar 8, 2005 #2

    dextercioby

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    What are the 2 equations...?Ang why is one part always negative...?

    Daniel.
     
  4. Mar 8, 2005 #3

    HallsofIvy

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    Have you considered that
    [tex]\frac{y^2}{b^2}- \frac{x^2}{a^2}= 1[/tex]
    is also an equation of a hyperbola?
     
  5. Mar 8, 2005 #4
    The equations are:

    [tex]b = \pm\sqrt{\frac{36a^2}{16 - a^2}}[/tex]
    [tex]b = \pm\sqrt{\frac{9a^2}{1 - a^2}}[/tex]

    Then when I try to solve for a after combining them, one side is always negative. I don't know why it is always negative, but I've tried solving for b instead of a first, and I encounter the same problem. I think someone else in my class had the same problem too. Maybe I just did it wrong or something..
    The teacher said something about subtracting those two equations once you get them..? How would you do that?
     
  6. Mar 8, 2005 #5

    HallsofIvy

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    Well, you just subtract! [itex]0= \sqrt{\frac{36a^2}{16-a^2}}- \sqrt{\frac{9a^2}{1-a^2}}[/itex] and now "move" that second term to the other side of equation: [itex]\sqrt{\frac{36a^2}{16-a^2}}= \sqrt{\frac{9a^2}{1-a^2}}[/itex] and now square both sides.

    Do you see where those formulas are from? If you assume a formula of the form [itex]\frac{x^2}{a^2}- \frac{y^2}{b^2}= 1[/itex], then putting x= 4, y= 6 gives [itex]\frac{16}{a^2}- \frac{36}{b^2}= 1[/itex] and putting x= 1, y= -3 gives [itex]\frac{1}{a^2}-\frac{9}{b^2}= 1[/itex]. Solve each of those for b and you get those two equations. Actually, what I would do is Multiply that second equation by 4 to bet [itex]\frac{4}{a^3}-\frac{36}{b^2}= 4[/itex]. Now subtract the first equation from that to eliminate b: [itex]\frac{12}{a^2}= -3[/itex] . Yes! You do have a square on one side equal to a negative number!

    Okay, so do what I suggested before: try writing your basic equation as
    [tex]\frac{y^2}{b^2}- \frac{x^2}{a^2}= 1[/tex]
    and do the same as before.
     
  7. Mar 8, 2005 #6
    Oh, I think I finally got the answer. I should've used [tex]\frac{y^2}{b^2}- \frac{x^2}{a^2}= 1[/tex] since the graph opens up and down. I could've sworn I did that when I attempted this the first time, but I couldn't get the answer for some reason.. Well, nevermind that. Thanks a lot for your help!
     
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