Proof of why conics map to conics after a perspective transformation

Question

Background

Consider a world where the ground is the standard $x$-$y$ plane with a Cartesian grid on it. The graph of a parabola $x^2 = 4ay$ is on this $x$-$y$ plane. A person of with eye-level $h$ above the ground is walking along the ground and stops a bit before the origin of the $x$-$y$ plane which is on the ground. He/she looks straight out in the direction of the positive $y$-axis and instead of a parabola sees an ellipse.

The following images and video animation illustrate this:

View of parabola straight down from above:

View of parabola with eyes at some height $h$ looking towards the horizon (positive $y$-axis):

Animation: https://www.youtube.com/watch?v=ukmqwGbfEZM

Question

My question is this:

Is there a way to prove that a general conic under the type of perspective projection I described here (where you go from looking straight down from above at the conic on the ground to looking straight out along the $y$-axis with your eyes at some height $h$ above the ground) maps to another conic (such as in this case from parabola to ellipse)?

More specifically, is there a way to convert the equation of a general conic in the first perspective (such as $x^2=4ay$) to an equation of the other conic in the second perspective given the parameters I described above (namely $h$ and that the person is looking out in the direction of the positive $y$-axis)?

Answer 1

Consider this diagram of viewing the $x$-$y$ plane from this perspective:

The eye/retina of the viewer is a length of $d$ away from the plane perpendicular to the plane of the ground and passing through the $x$-axis. This plane is the plane in which the $x$-$y$ plane is being viewed. The point is also a height $h$ above the ground. The viewer/viewing plane has a coordinate system $x'$-$y'$.

The point $O' = (0,0)$ for the $x'$-$y'$ coordinates corresponds to the point $(0, \infty)$ in the $x$-$y$ coordinates. All other points for the viewer are below the viewer's $x'$ axis.

Using similar triangles we get that $\frac{x'}{d} = \frac{x}{y+d}$ and $\frac{y'}{d} = -\frac{h}{y+d}$ (the $-$ sign is because everything is below the $x'$ axis for the viewer).

Rearranging these equations we get $y'(y+d)=dh$ so $y = \frac{-d(y'+h)}{y'}$ and $x'y+dx'=dx$ so $x = \frac{x'y}{d}+x'$. Plugging the expression for $y$ into the expression for $x$ we get $x = x'(1-\frac{(y'+h)}{y'}) = \frac{-hx'}{y'}$.

Thus, the final transformations are $x = \frac{-hx'}{y'}$ and $y = \frac{-d(y'+h)}{y'}$.

The equation of a general conic is $Ax^2+By^2+Cxy+Dx+Ey+F = 0$.

Plugging these transformations in we get:

$h^2(A+dC)x'^2+d(Bd-E+1)y'^2+h(dC-D)x'y'+Ch^2dx'+hd(2Bd-E)y'+Bh^2d^2 = 0$

This is an equation of the form $A'x'^2+B'y'^2+C'x'y'+D'x'+E'y'+F' = 0$, which is also the equation of a conic. Thus, conics are mapped to conics under this transformation.

For example, the parabola $y = x^2$ is mapped to the ellipse $\frac{x'^2}{1/4}+\frac{(y'+\frac{1}{4})^2}{1/16} = 1$ for $d = 1$ and $h = \frac{1}{2}$ as shown in the image below.

The parabola $y=x^2$ under the transformation for $d = 1$ and $h = \frac{1}{2}$:

Answer 2

The correct ambient space where to study these transformations is the projective plane $\mathbb{P}^2$ and they are called homographies.

E.g. see here

Homographies send collinear points into collinear points, thus transform lines into lines and thus they can be described analitically by linear equations in the homogeneous projective coordinates. This would follow also from the fact that homographies are isomorphisms of $\mathbb{P}^2$, in particular invertible.

But then the homographic transform of a curve of degree $d$ is a curve of degree $d$. In particular (when $d=2$) conics are tranformed into conics.