Invent transformation mapping ellipsoid to unit sphere Invent a transformation that maps the ellipsoid $ x^2+8y^2+6z^2+4xy-2xz+4yz=9$ onto the unit sphere. 
I don't even know where to begin with this question, any help would be appreciated.
 A: I always like to convert these quadratic equations into quadratic forms involving matrices, because then you can just find the eigenvalues and eigenvectors of the matrix to give you the transformation.
We want to write our system in the form
$$ \frac{1}{9}r^T A r = 1 $$
where $r = [x, y, z]^T$ and $A$ is a symmetric positive definite $3\times 3$ matrix.
If you expand the product out and match coefficients, you get
$$ \frac{1}{9}\begin{bmatrix}x\\y\\z\end{bmatrix}^T \begin{bmatrix}1 & 2 & -1 \\ 2 & 8 & 2 \\ -1 & 2 & 6\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} = 1 $$
Now, if you diagonalize the matrix $A$ so that $A = T^T \Lambda T$ (which is always possible since $A$ is symmetric positive definite), then you can let $p = \frac{1}{3}\sqrt{\Lambda}Tr$, so
$$ p^T p = 1 $$
This gives you the equation of the unit sphere in the coordinates of $p$. Therefore the transformation you want is $[x,y,z]^T \to \frac{1}{3}\sqrt{\Lambda}T [x,y,z]^T $.
Details: Since $A$ is a real symmetric matrix, $T$ is an orthogonal matrix, and $\Lambda$ is diagonal and real, so $\sqrt{\Lambda}$ just involves taking the square roots of all the diagonal elements. For this problem, with the numbers given, it appears that there are not "nice" values for these matrix entries.
A less "linear algebra"-y method
You can brute force this by working backwards. Start with a unit sphere $p^2 + q^2 + r^2 = 1$ in $(p,q,r)$ coordinates, and let each be linear combinations of $(x,y,z)$ coordinates:
$$ p = a_p x + b_p y + c_p z $$
$$ q = a_q x + b_q y + c_q z $$
$$ r = a_r x + b_r y + c_r z $$
Substitute these into the sphere equation, and expand. Then match terms with the original equation and solve for the 9 coefficients. This is exceedingly tedious. Note that you have 9 unknowns, and you only need to match 6 numbers. In the method I described above, there are 3 additional constraints arising from the fact that you need an orthogonal transformation (in other words, the new $p$ direction should be orthogonal to the $q$ and $r$ directions, etc.). However, if you just want some transformation, then that is not necessary, and you can match coefficients in whatever way is convenient.
A: The polynomial
$$f(x,y,z):=x^2+8y^2+6z^2+4xy-2xz+4yz=(x+2y-z)^2+4(y+z)^2 +z^2$$
is positive definite and assumes the value $9$ on your ellipsoid $E$. It follows that for each ${\bf u}\in S^2$ the function 
$$\phi_{\bf u}(t):=f(t{\bf u})\qquad(t\geq0)$$
is linearly increasing and assumes the value $9$ at the moment $t$ the ray $t\mapsto t{\bf u}$ is intersecting  $E$. Therefore the restriction of the map
$$(x,y,z)\mapsto{f(x,y,z)\over 9}{(x,y,z)\over \sqrt{x^2+y^2+z^2}}$$
(which scales each ray in its particular way) to $E$ will map $E$ bijectively onto $S^2$.
