I'm doing some work in projective geometry, and am given an example. I will type the full example out, and then the related problem and provided solution. Sorry in advance for the length of this!
Provided example:
In a three-dimensional rectangular coordinate system the viewing point, $C$, is the point $(0, -3, 2)$, the picture plane is the plane $y = 0$, and the plane containing the scene is the plane $z = 0$. What is the image on the picture plane of the point $(2, 1, 0)$? What is the image of the family of parallel lines $y = x + k$ in the scene? What is the image of the circle, $\Gamma$, whose equation in the plane of the scene is $x^2 + (y - 2)^2 = 1$?
It will be helpful in solving this problem to introduction two auxiliary coordinate systems in addition to the basic $x, y, z$ system itself. One of these, an $X$, $Y$ system in the $xy$ plane, we shall need in order to describe configurations which are limited to the object plane, $z = 0$. The other, an $X', Z'$ system in the $xz$ plane, we shall need in order to describe configurations which are limited to the image plane, $y = 0$.
By definition, the image of a point $P:(X, Y, 0)$ in the object plane is the point $P':(X', 0, Z')$ in which the line $PC$ intersects the image plane. Now we know from analytical geometry that the equations of the line determined by two points $(x_0, y_0, z_0)$ and $(x_1, y_1, z_1)$ can be written in the form $\dfrac{x - x_0}{x_1 - x_0} = \dfrac{y - y_0}{y_1 - y_0} = \dfrac{z - z_0}{z_1 - z_0}$. Hence, taking $C:(0, -3, 2)$ as the point $(x_0, y_0, z_0)$ and $P:(X, Y, 0)$ as the point $(x_1, y_1, z_1)$, we have for the equations of the projecting line $PC$, $\dfrac{x - 0}{X - 0} = \dfrac{y + 3}{Y + 3} = \dfrac{z - 2}{0 - 2}$. To find the coordinates $(X', 0, Z')$ of the point $P'$ in which this line pierces the picture plane, we merely put $y = 0$ and solve these equations for $x$ and $z$, getting
$$\tag{1} x \equiv X' = \dfrac{3X}{Y + 3} \ \ \ z \equiv Z' = \dfrac{2Y}{Y + 3}$$
To find the image of the given point $(2, 1, 0)$, we merely let $X = 2$, $Y = 1$ in the last pair of equations. The result is the image point $X' = \dfrac{3}{2}, Z' = \dfrac{1}{2}$, that is, the point $\left( \dfrac{3}{2}, 0, \dfrac{1}{2} \right)$.
To find the image of the family of parallel lines defined in the object plane by the equation $y = x + k$, that is, $Y = X + k$, it is convenient to solve Eqs. (1) for $X$ and $Y$, getting
$$\tag{2} X = \dfrac{2X'}{2 - Z'} \ \ \ Y = \dfrac{3Z'}{2 - Z'}$$
These equations, of course, express the coordinates of the object point $P:(X, Y, 0)$ in terms of the coordinates $(Z', 0, Z')$ of its image, $P'$. Substituting the expressions (2) into the equation $Y = X + k$, we find that in the picture plane the coordinates of the image point satisfy the equation
$$\dfrac{3Z'}{2 - Z'} = \dfrac{2X'}{2 - Z'} + k$$
or
$$(3 + k)Z' = 2X' + 2k$$
For all values of $k$ this line passes through the point $X' = 3, Z' = 2$, which is a point on the vanishing line $Z' = 2$, as required by theorem 2.
To find the image of the circle $\Gamma$, we merely substitute the expression (2) into the equation of $\Gamma$, namely $x^2 + (y - 2)^2 = 1$, that is, $X^2 + (Y - 2)^2 = 1$, getting $\left( \dfrac{2X'}{2 - Z'} \right)^2 + \left( \dfrac{3Z'}{2 - Z'} - 2 \right)^2 = 1$ or $(X')^2 + 6(Z')^2 - 9Z' + 3 = 0$. It is easy to see that this is not the equation of a circle but rather the equation of an ellipse.
Problem statement:
In example 1, what is the image of the parabola whose equation in the object plane is $Y = X^2$? What is the image of the parabola whose equation is $Y = 1 - X^2$?
Provided solution:
Ellipse $4(X')^2 + 3(Z' - 1)^2 - 3$. Hyperbola $2(X')^2 - 2(Z' - \frac{5}{4})^2 = -\frac{9}{8}$.
My question:
My first problem is that I don't understand how we do this exercise. As you can see, the provided solution is very brief, providing just the result without explanation. But my second problem is that I do not see how the solution provided is sensible. The problem asks us for parabolas, but the solution then provides an ellipse and a hyperbola. I understand that these are all conic sections, but how did we go from a parabola to an ellipse and hyperbola? And, more importantly, how do we even go about reasoning about such a problem and arriving at the solution?
I apologise for making this so lengthy. It would be great to get an explanation for how to do this problem, and how the author arrived at such an unintuitive solution. Thank you.
Let me explain this step by step.
The transformation is defined as $P\mapsto CP\cap\{ Y = 0 \}$, i.e. intersection of the line $CP$ with the $XZ$-plane.
For a point $P = (X, Y, 0)$, a line through $C = (0,-3,2)$ and $P$ is given by $$\{ t(P-C) + C\mid t\in\mathbb R\} = \{ (tX, t(Y+3)-3, -2t + 2) \mid t\in\mathbb R\}.$$
We need to find $t$ for which the corresponding point of the above line lies in the $XZ$-plane, so set $t(Y+3)-3 = 0$ to get $t = \frac 3{Y+3}$. Therefore, the intersection is $(\frac{3X}{Y+3},0,\frac{2Y}{Y+3})$, i.e. the transformation is $$(X,Y,0)\mapsto (\frac{3X}{Y+3},0,\frac{2Y}{Y+3}).$$
The above transformation can be written concisely as \begin{align} X' &= \frac{3X}{Y+3}, \\ Z' &= \frac{2Y}{Y+3}.\end{align}
Treat it as a system of equations and solve it for $X$ and $Y$ to get \begin{align} X &= \frac{2X'}{2-Z'}, \\ Y &= \frac{3Z'}{2-Z'}.\end{align}
In essence, this calculates the inverse transform $(X',0,Z')\mapsto (\frac{2X'}{2-Z'},\frac{3Z'}{2-Z'},0)$.
So, if you are given point $P = (2,1,0)$, set $X = 2$, $Y = 1$ to get $X' = 3/2$, $Z' = 1/2)$, i.e. $(2,1,0)\mapsto (3/2,0,1/2)$.
Roundabout way to do this is the following. We are looking at the set of points that satisfy $X = 2,\ Y = 1$. Substituting $X$ and $Y$ we get the set of all points that satisfy $\frac{2X'}{2-Z'} = 2$ and $\frac{3Z'}{2-Z'} = 1$. Solving gives us $X' = 3/2$ and $Z' = 1/2$.
In general, we are given the set of all points satisfying $f(X,Y) = 0$. To transform that curve, all you need to do is substitute $X$ and $Y$ to get the set of all points $f(\frac{2X'}{2-Z'}, \frac{3Z'}{2-Z'}) = 0$. Solving that equation gives us the curve that the original curve transforms to.
So, $$Y = X^2 \implies \frac{3Z'}{2-Z'} = \left(\frac{2X'}{2-Z'}\right)^2 \implies 4(X')^2 + 3(Z'-1)^2-3 = 0,$$ and similarly $$Y = 1-X^2 \implies \frac{3Z'}{2-Z'} = 1 - \left(\frac{2X'}{2-Z'}\right)^2 \implies 2(X')^2-2(Z'-5/4)^2=-9/8.$$
Imagine that you are standing at the point $C$ and tracing with your eyes the conic section $Y = X^2$. The rays that are formed between your eyes and the conic section form a cone. Now fix this cone. If you intersect it with the plane $Z = 0$, you get the beginning parabola. If you intersect it with some other plane, you'll get a different conic section. Our calculation shows that if you intersect it with the plane $Y = 0$, you get an ellipse.