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!
 A: 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 $\frac34y^2+z^2=1$.
A: In case you have difficulties with the accepted answer, here is another approach:
Your ellipsoid $S$ is  the level surface $F^{-1}(\{0\})$ of the function
$$F(x,y,z):=x^2+y^2+z^2-xy-1\ .$$
Let $S'$ be the shadow of $S$ in the $(x,y)$-plane. The points  $p\in S$ that generate the boundary $\partial S'$ are characterized by the property that the tangent plane at $p$ is vertical (i.e., parallel to the $z$-axis), which is the same as saying that the surface normal $n_p$ at $p$ is horizontal. Now $n_p$ is given by $\nabla F(p)$, whose third coordinate is $2z$.
It follows that the points $(x,y,z)\in S$ producing the shadow boundary are the points with $z=0$, i.e., the points $(x,y,0)$ satisfying $x^2+y^2-xy-1=0$. In this special example the shadow boundary in fact coincides with the intersection of $S$ with the $(x,y)$-plane.
Therefore let's do the shadow $S''$ of $S$ in the $(x,z)$-plane as well. The points  $p\in S$ that generate the boundary $\partial S''$ are characterized by the property that the tangent plane at $p$ is  parallel to the $y$-axis, which is the same as saying that the surface normal $n_p$ at $p$ is orthogonal to the $y$-axis, or has $y$-component $0$. The $y$-component of $\nabla F(p)$ is given by $2y-x$. Therefore we can say that the points $p$ in question lie on the plane $2y-x=0$. As they satisfy a priori $F(x,y,z)-1=0$ we can eliminate $y$ from the two equations
$$x^2+y^2+z^2-xy-1=0,\qquad y={x\over2}$$
and obtain a single equation connecting their $x$- and $z$-coordinates:
$$x^2+{x^2\over4}+z^2-x\>{x\over2}-1=0\ ,$$
or
$${3\over4}x^2+z^2-1=0\ .$$
This is already the equation of the shadow boundary $\partial S''$.
I may now safely leave the shadow on the $(y,z)$-plane to you.
A: By coordinate planes, I assume you mean the $xy$-plane, $xz$-plane, and $yz$-plane? If so, the projections can be found as follows:
$xy$-plane: Let $z=0$.
$xz$-plane and $yz$-plane: At first try, you may want to use $y=0$ or $z=0$ respectively; however, this will leave you with the cross section with the $xz$-plane, not the projection onto the $xz$-plane.  To get the proper projections, we need to analyze the graph of the projection in the $xy$-plane; in particular, find the equations of the horizontal and vertical tangent lines.
In the $xy$-plane, the equation simplifies to $x^2+y^2-xy-1=0$.  Implicitly differentiating with respect to $x$ gives us
$$2x+2y\frac{dy}{dx} - y - x\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{y-2x}{2y-x}.$$
Now, you get horizontal tangent lines when the numerator is zero (i.e. when $dy/dx = 0$), which implies that $y-2x=0\implies y=2x$.  Substituting this back into the implicit equation gives us 
$$x^2+4x^2-2x^2 -1 = 0 \implies 3x^2=1 \implies x=\pm \frac{\sqrt{3}}{3}$$
and thus $y = \pm\dfrac{2\sqrt{3}}{3}$ are the equations of the horizontal tangent lines.
Similarly, we find the vertical tangent lines when the denominator is zero (i.e. when $dy/dx$ is undefined), which implies that $2y-x=0\implies x=2y$.  In a similar fashion, we get that $y=\pm\dfrac{\sqrt{3}}{3}$ and hence $x=\pm\dfrac{2\sqrt{3}}{3}$ are the equations of the vertical tangent lines.

Now why did we go through all of this? Well, the equations of the tangent lines tells us how far along the $x$ and $y$ axis the ellipse extends; in particular, the vertical tangents give us a bound on $x$ and $y$ for the ellipse (i.e. $-2\sqrt{3}/3 \leq x,y \leq 2\sqrt{3}/3$). These bounds play the role of the major axes for the $xz$ and $yz$ projections of the ellipsoid onto the $y=0$ and $x=0$ planes respectively.  The minor axis will be along the $z$ direction; in particular, along the ellipsoid, $-1\leq z\leq 1$.  With that said, the corresponding projection of the ellipsoid onto the $yz$-plane is
$$\frac{3y^2}{4} + z^2 = 1$$

and the corresponding projection onto the $xz$-plane is
$$\frac{3x^2}{4} + z^2 = 1$$

Note that the last two figures are cylinders of the projections as seen in three space; the purpose of visualizing them this way was to show you that they completely enclose the ellipsoid of interest.
