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Equation of tangents of hyperbolas

  1. Feb 9, 2013 #1
    find the equation of the tangents to the hyperbola H` with equation [itex] \frac{x^2}{25} - \frac{y^2}{16} = 1 [/itex] at the point (1,4)

    in an earlier part of the equation we had to prove that a tangent to the a hyperbola in the form of [itex] \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 [/itex] is in the form of [itex] a^2m^2 = b^2 + c^2 [/itex] where the tangent is in the form of y = mx + c

    so I differentiated with respect to x and got

    [tex] \dfrac{dy}{dx} = \dfrac{16x}{25y} [/tex]

    subbing in the values of x and y I get the value of dy/dx to be 4/25

    dy/dx is the gradient of the tangent so subbing that into the equation [itex] a^2m^2 = b^2 + c^2 [/itex] as well as the values for a^2 and b^2 I can't get any real values for the constant, c so I'm not sure where I've gone wrong.
     
  2. jcsd
  3. Feb 9, 2013 #2

    SteamKing

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    You have calculated m from evaluating dy/dx for the hyperbola at (1,4)

    the equation of the tangent line is y = mx+c, so c (for the line) can be determined
    by subbing in (1,4) for (x,y)

    The relation a^2m^2 = b^2+c^2 is not that of a tangent line
     
  4. Feb 9, 2013 #3
    what is the relation of a^2m^2 = b^2+c^2 then?
     
  5. Feb 9, 2013 #4

    SteamKing

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  6. Feb 9, 2013 #5
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