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Homework Help: Hyperbola homework

  1. Jan 23, 2008 #1
    http://img184.imageshack.us/img184/6506/hmmyh4.jpg [Broken]

    The question ebfore had me find the equation of the tangent for any parametizaton values of (x(t),y(t)).

    Which is y = -x/t^2 + x(t)/t^2 + y(t)

    I'm pretty sure the case that a tangent to the hyperbola cant pass through the point is because the point lies inside the curve.

    So basically all i have to do is show that for x = 4 in the original equation has a y value less than 4, and for y = 4 in the equation has a x value less than 4? By the way I sketched the graph.

    Did I do this right?
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 23, 2008 #2


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    Homework Helper

    I don't understand your answer for the parametrisation of the hyperbola, the expression gives me y = y(t), which doesn't say much, to say the least.

    I don't think you have to parametrise the curve anyway. You first have to find dy/dx by differentiating implicitly and then suppose there is a tangent line to the curve which passes through (4,4) in the form y= mx + c. Let m=dy/dx. Just make sure you express dy/dx in terms of one variable, either y or x only.

    Then you have 2 equations:

    [tex]4 = 4m + c [/tex]

    [tex] y_{0} = mx_{0} + c[/tex]

    where [tex]y_{0} and x_{0}[/tex] is the point on the hyperbola for which a tangent line to it passes through (4,4). As above, you should be able to express [tex]y_{0}[/tex] in terms of [tex]x_{0}[/tex] alone. Then you now have 2 equations and 2 unknowns, x0 and c. Try solving for them and see what happens.

    I'm not sure what you're trying to achieve with your quasi-graphical method, being able to sketch the graph helps, but it's entirely possible to do this analytically as above.
    Last edited: Jan 23, 2008
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