Projection of ellipsoid Find the projections of the ellipsoid
$$ x^2 + y^2 + z^2 -xy -1 = 0$$
on the cordinates plan
I have no idea how to do this. I couldn't find much on google to help me with it too.
Thanks in advance!

An approach related to Christian Blatter’s answer can be found in Result 8.9 on page 201 of Hartley & Zisserman’s Multiple View Geometry In Computer Vision. Under the camera matrix $\mathtt P$, the outline of the quadric $\mathtt Q$ is the conic given by $\mathtt C^*=\mathtt P\mathtt Q^*\mathtt P^T$. (Here the superscript asterisk indicates the dual conic.) The proof follows directly from the observations that lines $\mathbf l$ tangent to the outline satisfy $\mathbf l^T\mathtt C^*\mathbf l=0$ and back-project to planes $\mathtt P^T\mathbf l$ that are tangent to the original quadric.  
In the present problem the quadric is the nondegenerate ellipsoid $$\mathtt Q = \begin{bmatrix}1&-\frac12&0&0 \\ -\frac12&1&0&0 \\ 0&0&1&0 \\ 0&0&0&-1\end{bmatrix}$$ so we can use inverse matrices for the duals: the outline of $\mathtt Q$ under the projection $\mathtt P$ is therefore $\mathtt C = (\mathtt P\mathtt Q^{-1}\mathtt P^T)^{-1}$.  
For orthogonal projection onto the $x$-$y$ plane, we can use $$\mathtt P = \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\end{bmatrix}$$ for the camera matrix and have $$\mathtt Q^{-1}=\begin{bmatrix}\frac43&\frac23&0&0\\\frac23&\frac43&0&0\\0&0&1&0\\0&0&0&-1\end{bmatrix},$$ which gives us $$\mathtt C = \begin{bmatrix}1&-\frac12&0\\-\frac12&1&0\\0&0&-1\end{bmatrix},$$ or $x^2+y^2-xy=1$. For the $x$-$z$ plane, $$\mathtt P = \begin{bmatrix}1&0&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}$$ and $$\mathtt C = \begin{bmatrix}\frac34&0&0\\0&1&0\\0&0&-1\end{bmatrix},$$ i.e., $\frac34x^2+z^2=1$. A similar computation for the $y$-$z$ plane yields $\frac34y^2+z^2=1$.
