To begin, I noted two things. Given how light works, the curve must be a radial projection (from the light source) of the bottom boundary of the shade (a circle) onto the wall. Second, only the half of the circle closest to the wall would actually contribute to the shadow on the wall, and thus the curve. And so begins the math:
In this particular waiting room, the shade was actually touching the wall. That is, there was a single point on the bottom boundary of the shade touching the wall, which was the point at the 'apex' of this curve where the shadow began. Therefore, I imagined the scenario as follows:
The bottom boundary of the shade can be represented as the unit circle $\alpha = \langle \cos(\theta) + 1, \sin(\theta), 0 \rangle$. The wall can be thought of as the $yz$-plane, and perhaps the light source was centered at $P = (1, 0, 1)$. Note that the half of the circle contributing to the shadow is $\frac{\pi}{2} \leq \theta \leq \frac{3\pi}{2}$.
Next, recall that the curve is a radial projection of the circle, so we first wish to find a parametrization of the line from $P$ to an arbitrary point on the circle $\alpha$. Such a parametrization is as follows: $l(t) = \langle 1 + \cos(\theta)t, \sin(\theta)t, 1-t \rangle$. Note that at $t = 0$, we are at the point $P$, and at $t = 1$, we are at the point $\langle 1 + \cos(\theta), \sin(\theta), 0 \rangle$ on the circle.
We want to know where this family of lines intersects the $yz$-plane. Well, the $x$-coordinate at the intersection points is $0$, so we get the following equations:
$$x = 1 + \cos(\theta)t = 0$$
$$y = \sin(\theta)t$$
$$z = 1 - t$$
Hence, $t = -\sec(\theta)$. Plugging this in, we get the following system of equations describing the curve in the $yz$-plane:
$$y = -\tan(\theta)$$
$$z = 1 + \sec(\theta)$$
For simplicity, we can just pretend we're back in the $xy$-plane, and $x = -\tan(\theta)$ and $y = 1 + \sec(\theta)$. From here, we can get out of a parametrized equation and into rectangular form by noting that $y^2 - x^2 - 2y = 0$.
Recall that a general equation $A_{xx}x^2 + 2A_{xy}xy + A_{yy}y^2 + 2B_xx + 2B_yy + C = 0$ for constants $A_{\circ \circ}$ and $B_{\circ}$ describes a hyperbola provided that $ \det \left[ \begin{array}{ c c } A_{xx} & A_{xy} \\ A_{xy} & A_{yy} \end{array} \right] < 0$. Well, in our case, $A_{xx} = -1$, $A_{xy} = 0$, and $A_{yy} = 1$.
Therefore, we conclude that the curve is a hyperbola!