Is this Linear Algebra Approach for Tangent Plane of an Ellipsoid Legal? In my vectorial calculus course, I was basically asked for finding a tangent plane for an ellipsoid parallel to a given plane by means of the gradient. Exactly the question was:

Find the values of m for which the plane $x - 2y -2z  + m = 0$ is
tangent to the ellipsoid $x^2 +4y^2 +16z^2 = 144$

I solve it using the gradient, but I thought it could be a more interesting way to solve it. Firstly, I suppose that if we were making the exact same problem with a sphere of radius 12, we could take advantage of its symmetry…

*

*Modify the scale of the coordinates to transform the ellipsoid into a sphere, by making the substitutions $x = x'$, $ y = \frac{1}{2} y'$, $z = \frac{1}{4}z'$. Hence I am working in a kind of $\rm I\!R^{'3}$ instead of $\rm I\!R^3$

*Apply the same exact substitution to the plane, in order to bring it to my new space, as well.

*Use the normalised vector normal to the plane, and find the scalar $\alpha$ for which the vector line intersects the surface of the sphere. Let's call it $v_1^{'}$. Technically speaking, that vector represents the point in which the the plane in $\rm I\!R^{'3}$ is tangent to the sphere.

*Apply a linear transformation T to $v_1^{'}$, so that $T:\rm I\!R^{'3} \rightarrow  \rm I\!R^3$ whose associate matrix is
\begin{bmatrix}
1 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 4
\end{bmatrix}
In order to bring my $v_1^{'}$ to the original space.

*Once I have my new vector $v_1$, It also touches the ellipsoid at it's surface. But is clear that is not normal to my original plane.

*However, as it is the point of hypothetical contact of the plane with the ellipsoid, I just calculate the norm of the projection of $v_1$ on the my normal vector $n$, and used it as the distance required for my plane to move in order to be tangent to the ellipsoid. Using the distance between planes equation, I found my final m and finished the exercise.

However, I'm not sure if it's completely legal. And if it was, is there any other application of this technique?, we can use if for more complex problems?
 A: Your argument appears to use the fact that an invertible affine transformation preserves planes and points of tangency, which is correct. (A proof is given below.) However, here is a way to simplify the argument. First one applies the transformation $$x'=x,y'=2y,z'=4z,~{\rm or~}<x,y,z>=\left<x',\frac 1 2 y',\frac 1 4 z'\right>.$$ Then the ellipsoid becomes the sphere $$x'^2+y'^2+z'^2=144$$ and the plane becomes $$x'-y'-\frac 1 2z'+m=0.$$ In the case for a sphere centered at the origin, the distance $d$ from the origin to the point of tangency (and the plane) must be the radius of the sphere. It follows that $$d=\frac{|m|}{\sqrt{1+1+1/4}}=12\Rightarrow m=\pm 18.$$ In particular, there is no need to solve the points of tangency on the sphere (although they are easily solved as $(8,-8,-4),(-8,8,4)$) and no need to map back the points of tangency on the ellipsoid (which are $(8,-4,-1)$ and $(-8,4,1)$).
Proof of the fact mentioned above:
It is well known and widely used (for example in the plane geometry) that affine transformations preserve tangency. In your case here, one can spell out the details. For example, let $X=AX'+b$ be the change of coordinates in matrix form, where $X=<x,y,z>^T,X'=<x',y',z'>^T,\det(A)\neq 0$. Let $g(X')=f(AX'+b)$. By the Chain Rule, it can be checked that $$\left<\frac{\partial g}{\partial x'},\frac{\partial g}{\partial y'},\frac{\partial g}{\partial z'}\right>=\left<\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right>A,$$ i.e. $$ (\nabla g)^T_{|(x_0',y_0',z_0')}=(\nabla f)^T_{|(x_0,y_0,z_0)}A,$$ where $X_0'=(x_0',y_0',z_0'),X_0=(x_0,y_0,z_0)$ are corresponding points on the surfaces $g(X')=0,f(X)=0$, respectively.
Now the tangent plane at $(x_0,y_0,z_0)$ for $f(X)=0$ with equation $$(\nabla f)^T_{|(x_0,y_0,z_0)}\cdot (X-X_0)=0$$ can be written as $$(\nabla f)^T_{|(x_0,y_0,z_0)}A\cdot (A^{-1}X-A^{-1}X_0)=0$$ which is equivalent to $$(\nabla f)^T_{|(x_0,y_0,z_0)}A\cdot ((A^{-1}X-A^{-1}b)-(A^{-1}X_0-A^{-1}b))=0,$$ i.e. $$(\nabla g)^T_{|(x_0',y_0',z_0')}\cdot(X'-X_0')=0.$$ This shows that the tangent plane is preserved under the affine transformation.
Remark. Note that in general affine transformations do not preserve angles. But the normality of the gradients with the tangent planes hold in the respective coordinates.
