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Tangents to parabola

  1. Sep 21, 2008 #1


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    1. The problem statement, all variables and given/known data

    I am given the parabola [tex]y=\frac{x^2}{2}[/tex]

    I need to find the equations of the 2 tangents to the parabola that pass through the point A(5,-2)

    2. Relevant equations


    [tex]a = \frac{1}{2}[/tex]
    therefore: tangents pass through the points [tex]P(p,\frac{p^{2}}{2})[/tex] and [tex]Q(q,\frac{q^{2}}{2})[/tex]


    3. The attempt at a solution

    I began to check if the given point was outside the parabola
    i.e. [tex]y_{1}<\frac{x^{2}_{1}}{2}[/tex]

    [tex]-2<\frac{5}{2}[/tex] therefore, the point lies outside the parabola and there are 2 lines that will pass through the point, and are a tangent to the parabola.

    [tex]y+2=m(x-5)[/tex] where there are 2 values of m, each intersecting the parabola only once. i.e. tangent to parabola.

    From here I am totally stumped. I can't use the 1st derivative as I don't know the x value for which the gradient will pass through the point.
    Any help would be much appreciated.
    Last edited: Sep 21, 2008
  2. jcsd
  3. Sep 21, 2008 #2
    This means that your equation of your line must pass through two given points.

    (5, -2) is one of them, so your equation [tex]y+2=m(x-5)[/tex] is a good place to start.

    First, what is m? Remember, m is your slope...

    Once you got that, now where does that second point have to be? If the line is tangent to the parabola with equation [tex] y=\frac{x^2}{2} [/tex]...
  4. Sep 21, 2008 #3


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    You want to find (x, y) such that y= m(x- 5)- 2 and y'(x)= m. You should certainly be able to find the derivative of y in terms of x. Putting that in for y' gives three equations for x, y, and m. this will, of course, reduce to a quadratic equation so you can expect two solutions: two tangents.
  5. Sep 21, 2008 #4


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    Thank you, I managed to solve it.
    Last edited: Sep 22, 2008
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