Proof of why conics map to conics after a perspective transformation

Consider this diagram of viewing the $x$-$y$ plane from this perspective: enter image description here

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 $h = \frac{1}{2}$: enter image description here

Tags:

Geometry

Projective Geometry

Conic Sections

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