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Thread: Parabola Parameters

  1. #1
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    Exclamation Parabola Parameters

    Hi

    Find the coordinates of 3 points on the parabola x^2 = 4y such that the normals through these 3 points pass through the point (-12,15).

    I have no clue whatsoever on how to do this - the previous questions were so different. Please, any help?
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post
    Hi

    Find the coordinates of 3 points on the parabola x^2 = 4y such that the normals through these 3 points pass through the point (-12,15).

    I have no clue whatsoever on how to do this - the previous questions were so different. Please, any help?
    any point on the parabola ... $\displaystyle \left(x, \frac{x^2}{4}\right)$

    normal slope ... $\displaystyle m = -\frac{2}{x}$

    using the point-slope form of a linear equation that passes through $\displaystyle (-12,15)$ ...

    $\displaystyle 15 - \frac{x^2}{4} = -\frac{2}{x}(-12 - x)$

    $\displaystyle 15 - \frac{x^2}{4} = \frac{24}{x} + 2$

    $\displaystyle 0 = \frac{x^2}{4} - 13 + \frac{24}{x}$

    $\displaystyle 0 = x^3 - 52x + 96$

    roots are ...

    $\displaystyle x = 2$ , $\displaystyle x = -8$ , $\displaystyle x = 6$
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  3. #3
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    Quote Originally Posted by xwrathbringerx View Post
    Hi

    Find the coordinates of 3 points on the parabola x^2 = 4y such that the normals through these 3 points pass through the point (-12,15).

    I have no clue whatsoever on how to do this - the previous questions were so different. Please, any help?
    A "normal" to the parabola is perpendicular to the tangent line at that point. The slope of the tangent line at $\displaystyle \left(x_0, \frac{1}{4}x_0^2)\right)$ is $\displaystyle \frac{1}{2}x_0$ so the slope of the normal line is $\displaystyle -\frac{2}{x_0}$. A normal line to the parabola at $\displaystyle \left(x_0, \frac{1}{4}x_0^2\right)$ is $\displaystyle y= -\frac{2}{x_0}\left(x- x_0)+ \frac{1}{4}x_0^3$. Put x= -12, y= 15 into that and solve for $\displaystyle x_0$.
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