# points on the surface

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• Oct 17th 2009, 09:44 PM
zpwnchen
points on the surface
Quote:

Find the points on the surface http://webwork.asu.edu/webwork2_file...9e2a58ead1.png at which the tangent plane is parallel to the plane http://webwork.asu.edu/webwork2_file...64227b3791.png.
How do I get started?
• Oct 17th 2009, 10:41 PM
redsoxfan325
Quote:

Originally Posted by zpwnchen
How do I get started?

Find the points on the surface http://webwork.asu.edu/webwork2_file...9e2a58ead1.png at which the tangent plane is parallel to the plane http://webwork.asu.edu/webwork2_file...64227b3791.png.

Parallel planes have the same normal vector (i.e. $\langle 3,3,-1\rangle$). The constant term is irrelevant.

The equation for the tangent plane is

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

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

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

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

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

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

So the two points are:

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