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Thread: points on the surface

  1. #1
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    points on the surface

    Find the points on the surface at which the tangent plane is parallel to the plane .
    How do I get started?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by zpwnchen View Post
    How do I get started?

    Find the points on the surface at which the tangent plane is parallel to the plane .
    Parallel planes have the same normal vector (i.e. $\displaystyle \langle 3,3,-1\rangle$). The constant term is irrelevant.

    The equation for the tangent plane is

    $\displaystyle F_x(x_0,y_0,z_0)(x-x_0)+F_y(x_0,y_0,z_0)(y-y_0)+F_z(x_0,y_0,z_0)(z-z_0)=0$

    $\displaystyle F_x=2x$
    $\displaystyle F_y=6y$
    $\displaystyle F_z=8z$

    So you want $\displaystyle 2x_0=3k$, $\displaystyle 6y_0=3k$, and $\displaystyle 8z_0=-k$ (because you have to take into account the normal vector can be scaled by any constant $\displaystyle k$).

    Spoiler:
    So the points are $\displaystyle \left(\frac{3k}{2},\frac{k}{2},-\frac{k}{8}\right)$.

    Now you need to solve for $\displaystyle k$ given the initial equation:

    $\displaystyle \frac{9k^2}{4}+3\frac{k^2}{4}+4\frac{k^2}{64}=1 \implies k=\pm\frac{4}{7}$

    So the two points are:

    $\displaystyle \left(\frac{6}{7},\frac{2}{7},-\frac{1}{14}\right)$ and $\displaystyle \left(-\frac{6}{7},-\frac{2}{7},\frac{1}{14}\right)$
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