We want to map $P = (x,y,z)^\top$ to $P'=(x',y',z')^\top$.
All rays go through $C = (0,0,-5)^\top = (0,0,-d)^\top$ and hit the plane $z = 0$.
(Large version here and here)
We have the line with intersection
$$
(0,0,-d)^\top + t ((x, y, z)^\top - (0,0,-d)^\top) = (x', y', 0)^\top \iff \\
(tx,ty,t(z+d) - d)^\top = (x', y', 0)^\top
$$
so we need
$$
t(z+d) -d = 0 \iff t = d/(z+d)
$$
This leads to
\begin{align}
P'
&= (x', y', z')^\top \\
&= (x', y', 0)^\top \\
&= \left( \frac{d}{z + d} x, \frac{d}{z + d} y, \frac{d}{z + d} (z+d) - d \right)^\top \\
&= \left( \frac{d}{z + d} x, \frac{d}{z + d} y, 0 \right)^\top \quad (*)
\end{align}
So far we are in agreement regarding $x'$ and $y'$.
We have difference in $z'$, which should be
$$
z' = \frac{1}{z/d + 1} (z+d) - d = \frac{d}{z + d} (z+d) - d = 0
$$
and $w'$ will be different as well, see below.
Using homogeneous coordinates we can write the transformation $(*)$ as
$$
p' = T p \iff \\
\begin{pmatrix}
x' \\
y' \\
z' \\
w'
\end{pmatrix}
=
\begin{pmatrix}
d & 0 & 0 & 0 \\
0 & d & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 1 & d
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
z \\
1
\end{pmatrix}
\quad (**)
$$
we get a homogeneous image vector
$$
p' = \left( d x, d y, 0, z + d \right)^\top
$$
which can be normalized to
$$
p' = \left( \frac{d}{z + d} x, \frac{d}{z + d} y, 0, 1 \right)^\top
$$
Finally one can apply the above transformation $(**)$ to $p = (
10, -20, -10, 1)^\top$.
This gives $p' = (50, -100, 0, -5)^\top$ which normalizes to
$p' = (-10, 20, 0, 1)^\top$ or $P'=(-10,20,0)^\top$, where the $x'$ value agrees with the 2D image view shown above.
The original ellipse in space can be specified parametrically by
$ p(t) = v_0 + v_1 \cos t + v_2 \sin t $
where $v_0, v_1 , v_2 \in \mathbb{R}^3$ are the center, and the two semi-axes, respectively.
In terms of the vector $w = [\cos t, \sin t, 1 ]^T$, we have
$ p(t) = V w $ with $ V = [v_1, v_2, v_0] $
Now the rays from the eyepoint $E$ to $p(t)$ can be expressed parametrically as
$ q(t) = E + s (p(t) - E) = E + s F w $
with $F = [v_1, v_2, v_0 - E] $
From which it follows that
$ q(t) - E = s F w $
i.e.
$w = \dfrac{1}{s} F^{-1} (q(t) - E)$
However $w = [\cos t, \sin t , 1 ]$ satisfies
$w^T Q_0 w = 0 $
where $Q_0 = \text{diag} \{1, 1, -1 \} $
Therefore, the rays $q(t) $ satisfy
$ \dfrac{1}{s^2} (q(t) - E)^T F^{-T} Q_0 F^{-1} (q(t) - E) = 0 $
We can assume that $s \ne 0$. And this the equation of an elliptical cone that is formed by the point $E$ and $p(t)$.
To complete the analysis, I will define the vector
$N = v_0 - E $
to be the vector connecting $E$ to $v_0$, and will take the projection plane
to have a normal vector $N$ and passing through $r_0 = E + \alpha N $ for some $\alpha \gt 0$. Therefore, the equation of the projection plane is
$ N^T (r - r_0) = 0 $
Next, we'll apply an affine transformation to the vector $q$ of the cone as follows
$q' = F^{-1} (q - E) $
It then follows that $q'^T Q_0 q' = 0 $
And this is the equation of a right circular cone with axis along the $z'$ axis with a semi-vertical angle of $45^\circ$. Applying the same transformation to the projection plane, results in
$ r = E + F (r') $
which we substitute into the equation of the plane, to obtain,
$ N^T (E + F r' - r_0) = 0 $
noting that $r_0 - E = \alpha N$, then,
$ N^T F r' = \alpha (N^T N) $
Which can be written as
$ (F^T N)^T (r') = \alpha (N^T N) $
This plane (in the transformed space) is intersecting the right circular cone (also in the same transformed space), and the intersection will be an ellipse if the angle between the normal to the plane (which is given by $F^T N $) and the $z'$ axis is less than $ 45^\circ$. Using the dot product between $[0, 0, 1]^T$ and $(F^T N)$
we get,
$ k \cdot F^T N = [0, 0, 1] F^T N = N^T N = \cos \theta | F^T N | $
But $|F^T N | = \sqrt{ (v_1^T N)^T + (v_2^T N) + (N^T N)^2 }$
Since we want $\theta \lt 45^\circ$ , then we want,
$\dfrac{ N^T N }{\sqrt{(v_1^T N)^2 + (v_2^T N)^2 + (N^T N)^2 } } \gt \dfrac{1}{\sqrt{2}} $
which upon manipulating the denominator becomes:
$\dfrac{ 1 }{\sqrt{ \dfrac{|v_1|^2}{N^T N} \cos^2 \theta_1 + \dfrac{|v_2|^2}{N^T N} \cos^2 \theta_2 + 1 } } \gt \dfrac{1}{\sqrt{2}} $
where $\theta_1$ is the angle between $v_1$ and $N$ and $\theta_2$ is the angle between $v_2$ and $N$
The last inequality is equivalent to
$|v_1|^2 \cos^2 \theta_1 + |v_2|^2 \cos^2 \theta_2 \lt N^T N $
And this can be achieved if the vector $N$ is large enough with respect to $v_1$ and $v_2$.
Assuming condition is satisfied, then the intersection curve will be an ellipse which can be computed, but for our purposes, we can just assume that it is
$q'(t) = q_0 + q_1 \cos t + q_2 \sin t $
For some $q_0, q_1, q_2$
At the final step, we will obtain $q$ from $q'$ using the fact that $ q = E + F q' $
Thus, the intersection curve is
$q(t) = E + F q_0 + F q_1 \cos t + F q_2 \sin t $
which is an ellipse.
Best Answer
We start with a point in the 3D model of the scene; this has 3D coordinates $(x,y,z)$ As you know, coordinates $(x,y,z)$ are projected to coordinates $(x_p,y_p,z_{vp})$, which are coordinates in the view plane.
So already, because the view plane is a plane, everything has been reduced to two dimensions. But to transfer it to the computer screen, we recognize that the $z_{vp}$ is the same at all points in the view plane. Comparing two projected points, only $x_p$ and/or $y_p$ can be different.
From $x_p$ we get the horizontal coordinate on the computer screen. From $y_p$ we get the vertical coordinate on the computer screen. These two coordinates are enough to select the correct pixel on the 2D computer screen where the projected point should be displayed.
Yes, the projectors converge to the center of the projection (which must be at a non-zero $z$ coordinate).
No, the perspective projection is not reduced to parallel projection just by moving the view plane to $z_{vp}=0$. Perspective projection remains perspective projection no matter where you put the view plane. (Just keep the center the projection off the view plane, because if you put the center on the view plane you don't get a perspective or a parallel projection; everything just collapses onto the center point and you get no picture.)
Nobody uses four coordinates for a projective plane. The so-called points at infinity are written $(x, y, 0).$
The projective plane is a mathematical extension of the ordinary Euclidean plane. It's connected to the theory of perspective projection but you don't need to use the projective plane as a tool to perform any of the standard projections of computer models to images (and I do not see any way in which you would want to use it).
It is certainly possible mathematically to construct a perspective projection of a projective plane. Points at infinity would be projected at vanishing points in the projection. But this has nothing to do with the standard methods you will find when you look in textbooks and lectures on how perspective projection is done in software. Looking at the references posted in your earlier questions, for example, no projective plane is used in any way anywhere in them.
(Community wiki because I feel it would be unethical to take credit for answering a question that was directly caused by my own answers to previous questions.)