Find the point of tangency between a plane and an ellipsoid So, it is given that -
The tangent plane to the ellipsoid $4x^2 + y^2 + 2z^2 = 16$ is $2x + y + 2z = k$.
I’m trying to find k, and the point of tangency between those two.
What I did -
Assumed that the normal to the plane is parallel to the gradient of the ellipsoid, so I get $(8x, 2y, 4z) = (2, 1, 2)$. But then I get point $(x, y, z)$ which isn’t on the ellipsoid. 
Why is my assumption wrong?
 A: You’ve set the two vectors equal to each other, which is too strong. If they’re parallel, then all you can say for sure is that one is a scalar multiple of the other, i.e., you need $(8x,2y,4z)=\lambda(2,1,2)$ for some $\lambda\ne0$ instead. Since you’re working in three dimensions, you can avoid introducing an extraneous variable by stating the condition as $(8x,2y,4z)\times(2,1,2)=0$ instead. This will give you three linear equations, two of which are independent.
A: Substituting
the plane equation into the ellipsoid we have
$$
4 x^2 + (k - 2 x - 2 z)^2 + 2 z^2 = 16
$$
Solving for $x$ we have
$$
x = \frac{1}{4} \left(k-2 z\pm\sqrt{4 k z-8 z^2-k^2+32}\right)
$$
but at tangency we need
$$
4 k z-8 z^2-k^2+32 = 0
$$
solving now for $z$
$$
z = \frac{1}{4} \left(k\pm \sqrt{64-k^2}\right)
$$
but at tangency $64-k^2 = 0$ then $k = \pm 8$

NOTE
At tangency we have
$$
(8x,2y,4z)=\lambda(2,1,2)
$$
then 
$$
x = \frac{\lambda}{4}\\
y = \frac{\lambda}{2}\\
z = \frac{\lambda}{2}
$$
now after substitution into the ellipsoid equation we have the $\lambda$ value. Now with the tangency point, after substitution into the plane equation we get the $k$ value
A: The vector $(8x,2y,4z)$ is parallel to the vector $(2,1,2)$, not equal to it. 
$8x=4z\implies z=2x, 2\times2y=8x\implies y=2x$
Substitute for $y,z$ in $4x^2+y^2+2z^2=16$
$\implies 4x^2+4x^2+8x^2=16x^2=16$
$\implies x=\pm1$
$(x,y,z)\equiv\pm(1,2,2)$
Depending on the value of $k$, you will have to select one solution.
