how to get the equation of a surface equation like $F(x,y,z)=0$ by parallel projecting onto a plane? taking example 1:
we want to project the  standard  ellipsoid  equation:
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1
$$
onto the plane equation $$z=0$$
here we can just substitute $z$ with $0$
and get the projected equations:
$$
\begin{cases}
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\
z=0
\end{cases}
$$

taking example 2:
there is a non-standard  ellipsoid equation with center shifted:
$$
\frac{(x-u)^2}{a^2}+\frac{(y-v)^2}{b^2}+\frac{(z-w)^2}{c^2}=1
$$
too, project it onto  the plane equation $$z=0$$
if we just substitute $z$ with $0$ again.
we get the equation 1 :
$$
\begin{cases}
\frac{(x-u)^2}{a^2}+\frac{(y-v)^2}{b^2}=1-\frac{(w)^2}{c^2} \\
z=0
\end{cases}
$$
but by geometrical intuition , the equation 1 is not the corrected solution.
it is just a slice of the ellisoid cut by $z=0$
instead, by intuition, the projected equation should be the equation 2:
$$
\begin{cases}
\frac{(x-u)^2}{a^2}+\frac{(y-v)^2}{b^2}=1 \\
z=0
\end{cases}
$$
equation2 means that we can substitute $\frac{(z-w)^2}{c^2}$ with $0$ and get the projected equation.
but why this operating works? 
I don't know.

if we get a non-standard ellipsoid equation that not only center shifted but also center rotated.
how to get the corrected projected equation onto $z=0$?
like example3:
$$
2x^2+2y^2+2z^2-2(x+y+z)+2xy+2zy+2zx=0
$$
I don't know.

further more, a general surface or plane equation like $F(x,y,z)=0$, projecting it onto a general plane like $Ax+By+Cz+D=0$, how to get the corresponding projected equation? do we have the general method to get it?
 A: If you project a surface $S$ described in however way to the plane $z=0$ the "equation" of the projected surface $\pi_z(S)$  is $z=0$. But of course this is not what you want. If $S$ is an  ellipsoid surface  in space then $\pi_z(S)$ is a doubly covered elliptical disc $E$ in the $(x,y)$-plane. What you want to know is the exact boundary $\partial E$.
If $S$ is axis aligned and centered at ${\bf 0}$ then it so happens that you can just put $z=0$ in the equation of $S$, and a similar reasoning works if $S$ is still axis aligned, but offset. The following argument shows why this works in these special cases.
For the general case we have to find out how the boundary $\partial E$ comes about. Now $\partial E$ is the image of the points on $S$ where the tangent plane is vertical, i.e. parallel to the $z$-axis. Therefore we have to find these points. If $S$ is given by a quadratic polynomial equation $F(x,y,z)=0$ then at each point $(x,y,z)\in S$ the vector $\nabla F(x,y,z)$ points into the normal direction of $S$. This normal direction is horizontal (hence the tangent plane vertical) iff the third component ${\partial F\over\partial z}$ of $\nabla F(x,y,z)$ vanishes. It follows that the outline points of $S$ with respect to $\pi_z$ are satisfying the two equations
$$F(x,y,z)=0,\qquad{\partial F\over\partial z}(x,y,z)=0\ .\tag{1}$$
If $(x,y,z)$ satisfies $(1)$ then $\pi_z(x,y,z)=(x,y,0)$ is a point of $\partial E$. This fact can be cast into the following recipe: If you eliminate $z$ from $(1)$ you obtain the equation of $\partial E$ in the form $g(x,y)=0$.
A: The problem you are addressing is the difference between a projection and an intersection.
The intersection between two geometric objects given by set of points $(x,y,z)$ satisfying $P(x,y,z)=0$ for the first and $Q(x,y,z)=0$ for the second is the set of points $(x',y',z')$ satisfying $P(x',y',z')=0 \wedge Q(x',y',z')=0$.
The projection of a point $M$ of a geometric object given by set of points $(x,y,z)$ satisying $P(x,y,z)=0$ onto a plane is deduced by performing the following:


*

*Consider the line carrying the normal vector of the plane that passes through $M$

*The projection of $M$ is the intersection of the considered line with the plane 
The projection of a geometric object given by set of points $(x,y,z)$ satisying $P(x,y,z)=0$ onto a plane is the set of projections of every point of the geometric object.
The projection of a line to a plane is the line itself, the intersection of a line with a plane is a point or the empty set. 
For an ellipsoid, the intersection of the ellipsoid given by $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ onto the (XY) plane given by $z=0$ is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. The projection is given by $\frac{x^2}{a^2}+\frac{y^2}{b^2}\leq1$. Generally speaking the projection and intersection of two objects are two different things.
A: Simply substituting $0$ for $z$ to find the boundary of the first ellipsoid’s projection happens to work because the stars are favorably aligned. As others have pointed out, what you’re actually doing is computing the intersection of this surface with the plane $z=0$, which in this case coincides with the boundary curve of the projection. However, this doesn’t work if you translate the ellipsoid, as you’ve discovered. You can even restrict the translation to movement parallel to the $z$ direction, which is equivalent to projecting the original ellipsoid onto the plane $z=w$. As you’ve found, this boundary curve can be obtained by replacing $(z-w)^2/c^2$ by zero in the ellipsoid’s equation. Another way to achieve the same result is by setting $z=w$. This is suggestive, but still doesn’t quite explain why either of the substitutions works.   
In general, the image outline of a smooth surface $S$ under a central projection onto a plane results from points on $S$ at which rays from the projection center (viewpoint) $\mathbf V$ are tangent to $S$. These tangent rays form a cone (or cylinder in the case of parallel projection) and the image outline is the intersection of this cone with the image plane. The set of points on $S$ at which we have these tangent rays is called the contour generator $\mathbf\Gamma$ and the image of $\mathbf\Gamma$ under the projection is the apparent contour $\gamma$. The apparent contour is also called the “outline” or “profile.”  
For a quadric surface such as your ellipsoids, the apparent contour is a conic section. This is clear in the case of a sphere, since the apparent contour is the intersection of the image plane with a tangent cone to the sphere. The sphere can be mapped to another quadric via a 3-space projective transformation, which transforms the apparent contour to a conic, and since this transformation preserves intersection and tangency, the contour generator is also a plane conic.  
Switching now to homogeneous coordinates for convenience, we can quite easily find the apparent contour $\mathtt C$ of a quadric using dual quadrics and conics. Let $\mathtt Q$ be the matrix of the quadric and $\mathtt P$ the matrix of a central projection map from $\mathbb {RP}^3$ to $\mathbb {RP}^2$ (the camera matrix, in computer vision terminology). Lines $\mathbf l$ tangent to $\mathtt C$ satisfy $\mathbf l^T\mathtt C^*\mathbf l=0$, where $\mathtt C^*$ is the dual conic matrix, which for a nondegenerate conic is simply a nonzero multiple of $\mathtt C^{-1}$. These lines back-project to planes $\mathbf\pi=\mathtt P^T\mathbf l$ that are tangent to $\mathtt Q$, so satisfy $\mathbf\pi^T\mathtt Q^*\mathbf\pi=0$. We then have $$\mathbf\pi^T\mathtt Q^*\mathbf\pi = \mathbf l^T\mathtt P\mathtt Q^*\mathtt P^T\mathbf l = \mathbf l^T\mathtt C^*\mathbf l = 0,$$ and since this holds for all $\mathbf l$ tangent to $\mathtt C$, this means that $$\mathtt C^* = \mathtt P \mathtt Q^* \mathtt P^T. \tag1$$ The plane $\mathbf\pi_{\mathbf\Gamma}$ of the contour generator is just the polar plane of the projection center, $\mathtt Q\mathbf V$. For a parallel projection of a central quadric, this plane passes through the quadric’s center: $$(\mathtt Q\mathbf V)^T(\mathtt Q^*\mathbf\pi_\infty) = \mathbf\pi_\infty^T\mathtt Q^*\mathtt Q^*\mathbf V = \lambda \mathbf V^T \mathbf\pi_\infty$$ for some $\lambda\ne0$, but $\mathbf\pi_\infty=(0,0,0,1)^T$ and $\mathbf V$ is a point at infinity for a parallel projection, so this product vanishes. If the image plane is orthogonal to the projection direction, this means that you can find the apparent image of $\mathtt Q$, up to an orientation-preserving isometry, by intersecting the quadric surface $\mathtt Q$ and $\mathtt Q\mathbf V$. If the projection direction is parallel to a principal axis, then this plane is defined by the other two principal axes and the center of the quadric.  
We can now see why your substitutions in the first two examples worked. For projection parallel to the $z$-axis onto the $x$-$y$ plane, we have $\mathbf V=(0,0,1,0)^T$. For the first ellipsoid, $\mathbf\pi_{\mathbf\Gamma}$ is the $x$-$y$ plane itself, so its apparent image coincides with its intersection with this plane. For the translated ellipsoid, we have $\mathtt Q\mathbf V = (0,0,1/c^2,-w/c^2)^T$, i.e., the plane $z=w$. The projection of the resulting curve in this plane onto the $x$-$y$ plane is then obtained by setting $z=w$ in the ellipsoid equation.  
For your last example, $$\mathtt Q = \begin{bmatrix}2&1&1&-1\\1&2&1&-1\\1&1&2&-1\\-1&-1&-1&0\end{bmatrix}$$ and so $\mathbf\pi_{\mathbf\Gamma} = (1,1,2,-1)^T$. This plane is no longer parallel to the image plane, so the image of its intersection with the ellipsoid isn’t quite so simple to compute. We can use formula (1), which will involve two matrix inversions, but since the intersection of the tangent cylinder with our image plane is obtained by simply setting $z=0$, there’s an easier way to compute the apparent image. The tangent cone (cylinder) to the quadric is $$(\mathbf V^T\mathtt Q\mathbf V)\mathtt Q - (\mathtt Q\mathbf V)(\mathtt Q\mathbf V)^T.$$ For our projection, $\mathbf V^T\mathtt Q\mathbf V$ is just the $(3,3)$ element of $\mathtt Q$, and setting $z=0$ amounts to deleting the third row and column of the resulting matrices. We thus have $$2\begin{bmatrix}2&1&-1\\1&2&-1\\-1&-1&0\end{bmatrix} - \begin{bmatrix}1&1&-1\end{bmatrix}\begin{bmatrix}1\\1\\-1\end{bmatrix} = \begin{bmatrix}3&1&-1\\1&3&-1\\-1&-1&-1\end{bmatrix},$$ which corresponds to the implicit Cartesian equation $$3x^2+2xy+3y^2-2x-2y=1.$$
