What I am trying to find is a formula relating the projected area $A_p$ to the area $A$ of a planar object of arbirtrary shape, position and orientation in 3D space when subject to a projective transformation $P$ to 2D space, given the object's position and its orientation (e.g. by a plane normal) in relation to the origin. I would like to find this formula for an infinitesimally small shape, so that it becomes shape-independent and relates perceived and real area of any point in 3D space given its "orientation" (if you can say that about a point).
How I have approached this so far is by imagining a circle in 3D space, described by a centre point and two points on its circumference who I assume to be lying on the end of the two half-axes described by the ellipse created after projective transformation and then relating the area of the circle and the ellipse, given the radius, centre point and normal of the circle. However, I can't quite get it infinitesimally small nor do I know if I can just define my two points on the circumference to be lying on the end of the two half-axes of the ellipse. Is there a better approach? Does such a closed solution even exist and how does it look like?
Draw a small circle centered at $\vec P$, in the plane defined by the orthonormal vectors $\vec u$ and $\vec v$. A point of the circle is
$$\vec P+r\cos\theta\,\vec u+r\sin\theta\,\vec v.$$
With a focal distance $f$ and the projection with the pole at the origin onto a plane parallel to $xy$,
$$\begin{cases}X=f\dfrac{P_x+r\cos\theta\,u_x+r\sin\theta\,v_x}{P_z+r\cos\theta\,u_z+r\sin\theta\,v_z}, \\Y=f\dfrac{P_y+r\cos\theta\,u_y+r\sin\theta\,v_y}{P_z+r\cos\theta\,u_z+r\sin\theta\,v_z}.\end{cases}$$
Now using the Taylor develoment to the first order, we can approximate with
$$\begin{cases}X=\dfrac f{P_z}\left(P_x+r\cos\theta\,(u_x-u_z\dfrac{P_x}{P_z})+r\sin\theta\,(v_x-v_z\dfrac{P_x}{P_z})\right), \\Y=\dfrac f{P_z}\left(P_y+r\cos\theta\,(u_y-u_z\dfrac{P_y}{P_z})+r\sin\theta\,(v_y-v_z\dfrac{P_y}{P_z})\right).\end{cases}$$
The variable terms take the form
$$\begin{cases}\tilde X=ra\cos\theta+rb\sin\theta,\\\tilde Y=rc\cos\theta+rd\sin\theta,\end{cases}$$ which describe an ellipse. We get the area by
$$\int_0^{2\pi}\frac{X\dot Y-Y\dot X}2\,dt=\int_0^{2\pi}\frac{ad-bc}2\,dt.$$
Hence the requested ratio is
$$ad-bc$$ where these coefficients are as given above.