Tangent plane of a surface (an ellipsoid?) parallel to another plane I need to find the tangent plane of $x^2+2y^2+3z^2=21$ that is parallel to $x+4y+6z=0$.
Since our current topic covers parametrisations I was wondering if there is a good way to describe the curve in question as $\vec r(t)$.
Both paths to the solution would be appreciated.
 A: If we put $F(x, y, z) = x^{2} + 2y^{2} + 3z^{2}$, the gradient vector
$$
\nabla F(x_{0}, y_{0}, z_{0}) = (2x_{0}, 4y_{0}, 6z_{0})
$$
is normal to the level set $F(x, y, z) = 21$ at the point $(x_{0}, y_{0}, z_{0})$. It follows that the tangent plane to the ellipsoid is parallel to $x + 4y + 6z = 0$ if and only if
$$
(2x_{0}, 4y_{0}, 6z_{0}) \propto (1, 4, 6)
$$
and
$$
x_{0}^{2} + 2y_{0}^{2} + 3z_{0}^{2} = 21.
$$
(This approach is simple and workable, but has nothing to do with the techniques you mention.)
A: The the ellipsoid is a level surface of the function
$$F(x,y,z)=x^2+2y^2+3z^2 $$
so the normal vector at a point $(x_0,y_0,z_0)$ is equal to the gradient 
$$ \nabla F(x_0,y_0,z_0)=2x_0\mathbf{i}+4y_0\mathbf{j}+6z_0\mathbf{k} $$.
which must be parallel to the normal vector of the plane, so we have
$$ 2x_0\mathbf{i}+4y_0\mathbf{j}+6z_0\mathbf{k}=2t\mathbf{i}+4t\mathbf{j}+6t\mathbf{k} $$
for some $t$. Substituting $x_0=\frac{t}{2}$, $y_0=t$, $z_0=t$ into the equation of the ellipsoid and solving for $t$ yields $t=\pm2$.
So there are two points of tangency where the tangent plane is perpendicular to the line $x+4y+6z=0$, $(1,2,2)$ and $(-1,-1,-2)$.
\begin{eqnarray}
 2(x-1)+4(y-2)+6(z-2)&=&0\\
 2(x+1)+4(y+2)+6(z+2)&=&0\\
\end{eqnarray}
which resolve to
\begin{equation}
x+4y+6x\pm21=0
\end{equation}
