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Math Help - Parabola QUestion

  1. #1
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    Parabola QUestion

    The question goes:

    Find the y-intercept of the normal to the parabola y=x^2 at the point
    (a,a^2), where a cannot equal 0. Find the limit of this y-interept as a approaches 0. For each b>0 determine the number of normals that can be drawn from the point (0,b) to the parabola y=x^2; justify your answer.

    So far i have gotton that the y intercepts= 1/2+a^2, however, i'm not sure what the limit of the y intercept as a approaches 0. Additionally, i can't figure out how to determine the amount of normals.

    Thanks a Lot

    Willie Li
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  2. #2
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    Hello, aussiekid90!

    Find the y-intercept of the normal to the parabola y = x^2
    at the point (a,a^2), where a \neq 0

    (a) Find the limit of this y-intercept as a \to 0.

    You did the hard work . . . You found that the y-intercept is: . \frac{1}{2} + a^2

    As a\to0, we see that \left(\frac{1}{2} + a^2\right) \to \frac{1}{2}



    (b) For each b>0, determine the number of normals that can be drawn
    from the point (0,b) to the parabola y=x^2.

    We already know that the y-intercept is: . b \:=\:a^2 + \frac{1}{2}
    . . Then:. a \:=\:\pm\sqrt{b - \frac{1}{2}}

    In general, for every b, there are two values of a.


    Therefore, there are two normals that can be drawn . . .
    except: when b = \frac{1}{2}, then one normal can be drawn
    . . and when b < \frac{1}{2} when there are no normals.

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