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Math Help - Hyperbolas

  1. #1
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    Hyperbolas

    I'm not sure whether this should go on basic algebra or here, but since I'm learning this in college I'm gonna post it here. Ok, I'm in dire need of help with two problems:

    1) A hyperbola goes through the point P(6, 2), and one of its asymptotes is the line r: 2x + 5y = 0. Determine its equation.

    2) Prove that a line parallel to one asymptote of a hyperbola interesects it in a single point.

    I spent almost two hours on that first problem and tried everything I could, it sounds so fing simple yet I couldn't figure it out. Everything I tried filled up 3-4 pages of my notebook. I must not have understood some basic concept. I'm not gonna transcribe everything here because I think it would be pointless, but please believe me when I said I tried it until I couldn't stand it anymore. If you guys could just point me in the right direction...

    For the second one I equaled y = sqrt(b^2 * (x^2/a^2 - 1)) which I got from the reduced equation of a hyperbola to y = bx/a + d). I ended up with (b^2 - b)*x^2 + (2dba)*x + (d^2*a^2 + a^2*b^2). Then I supposed B^2 - 4AC should equal 0, which would mean it had only one solution but I just got a really ugly equation and couldn't see why it would equal 0.
    Last edited by topsquark; May 12th 2011 at 01:14 PM. Reason: Begging
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  2. #2
    Member HappyJoe's Avatar
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    For the first part (and probably also for the second part), it will help you to show that if \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 is the expression for the hyperbola, then the slopes of the asymptotes are \pm \frac{b}{a}, so that the equations of the asymptotes are y=\pm \frac{b}{a}x.

    Why? Solve the equation for the hyperbola with respect to y to get y=\pm b\sqrt{\frac{x^2}{a^2}+1}, and check that as x increases, the right hand side approaches y=\pm \frac{b}{a}x.

    The slope of your given asymptote is -2/5, so b/a = 2/5. Using the point P(6,2) as well gives you two equations with two unknowns.
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  3. #3
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    Quote Originally Posted by HappyJoe View Post
    For the first part (and probably also for the second part), it will help you to show that if \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 is the expression for the hyperbola, then the slopes of the asymptotes are \pm \frac{b}{a}, so that the equations of the asymptotes are y=\pm \frac{b}{a}x.
    But how do I do that? Would I have to use limits?

    Quote Originally Posted by HappyJoe View Post
    Why? Solve the equation for the hyperbola with respect to y to get y=\pm b\sqrt{\frac{x^2}{a^2}+1}, and check that as x increases, the right hand side approaches y=\pm \frac{b}{a}x.
    Sorry, I don't get how that would help me.

    Quote Originally Posted by HappyJoe View Post
    The slope of your given asymptote is -2/5, so b/a = 2/5. Using the point P(6,2) as well gives you two equations with two unknowns.
    So I have the equation from the quotation above, and also 2a=5b, right? So plugging in 6 for x and 2 for y on the above and substituting a for 5b/2, I end up getting {b}^{2 } - 4b +  - \frac{144}{25{b}^{2}} + 5 = 0 but then I'm stuck.
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  4. #4
    Member HappyJoe's Avatar
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    Quote Originally Posted by zehgs View Post
    But how do I do that? Would I have to use limits?
    Yeah.


    Quote Originally Posted by zehgs View Post
    Sorry, I don't get how that would help me.
    How do you mean? It helps you to show that the asymptotes have that particular slope, which will express a in terms of b.

    Edit: Oh, and it should be -1 inside of the square root instead of +1.

    Quote Originally Posted by zehgs View Post
    So I have the equation from the quotation above, and also 2a=5b, right? So plugging in 6 for x and 2 for y on the above and substituting a for 5b/2, I end up getting {b}^{2 } - 4b +  - \frac{144}{25{b}^{2}} + 5 = 0 but then I'm stuck.
    I'm not sure what you did to your equation, but it does look nasty. Substituting 5b/2 for a in \frac{36}{a^2} - \frac{4}{b^2} = 1 gives

    \frac{36}{\frac{25}{4}b^2} - \frac{4}{b^2} = 1,

    where I can put the left hand side on the common denominator \frac{25}{4}b^2.
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  5. #5
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    Yay, got it. Thank you very much.
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