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Homework Help: Parabola Find a the value of a

  1. Aug 7, 2012 #1
    The point P(2, 8) lies on the parabola C with equation y2=4ax. Find

    a the value of a,

    b an equation of the tangent to C at P.

    the value of a is 8, so y^2 = 32x

    when finding the tangent [itex] y = 4\sqrt{2} x^{\frac{1}{2}} [/itex] so at P [itex] \frac{dy}{dx} = \frac{2\sqrt{2}}{\sqrt{2}} [/itex] so the gradient of the tangent is 2, but why isn't it also -2? as when differentiating y^2 = 32x, you get also [itex] y = -4\sqrt{2} x^{\frac{1}{2}} [/itex] Could anyone explain why they have only used the positive value?
    Last edited by a moderator: Feb 4, 2013
  2. jcsd
  3. Aug 7, 2012 #2


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    hi synkk! :smile:

    (try using the X2 button just above the Reply box :wink:)

    i'm not really following you :confused:

    from y2 = 32x, for any positive x there are two values of y

    P is (2,8) and let's call Q (2,-8), so both P and Q lie on the parabola

    for P, the slope is 2, for Q it is -2 … where's the difficulty?

    (btw, it's a lot easier to use implicit differentiation … 2ydy/dx = 32, which works easily for either value :wink:)​
  4. Aug 7, 2012 #3
    Re: parabola

    Just my stupidity really,

    I was just wondering why in the solutions they didn't include -2, then you pointed out it was the other point, doh.

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