Today, I encountered a rather interesting problem in a waiting room:
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Notice how the light is being cast on the wall? There is a curve that defines the boundary between light and shadow. In my response below, I will prove, algebraically, that this curve is a hyperbola. I'm also interested in seeing a variety of other solutions, so please feel free to post your own.
Oh... this question was my homework a year ago, and this is my proof which is short as per my laziness.
The bad grey drawing is the locus required. Obviously, for this part the light rays just touch the plate. Now $$\dfrac{PA}{PD}=\dfrac{CA}{CB}=\text{constant}>1$$
Hence, the locus is a hyperbola as ratio of distance from the top of candle and line of candle(extended) is ...
This case is similar to the lamp except the fact that there is darkness below and light above the boundary as contrary to a lamp. But this has no effect on the boundary.
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!
This is a textbook case of conic sections, I might have even seen a lamp in the actual high school textbook. It's also very obviously a hyperbola because it has two branches. This helps you to see the solution without any computation!
You might argue that having both top and both illuminated parts doesn't prove hyperbola, because the geometry of the light might be different and so it wouldn't mean the parts belong to the same mathematical curve. However, you can always imagine the top lamp to be symmetric in such a way, that the double cone of light really is a double cone (with the same angle and same orientation in both directions). That way even if the physical light is not symmetric, each branch is still a parabola, even if not from the same curve.
