# Thread: points on the surface

1. ## 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?

2. Originally Posted by zpwnchen 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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