Today, I encountered a rather interesting problem in a waiting room:

$\qquad \qquad \qquad \qquad$enter image description here

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.

  • 5
    $\begingroup$ Kaj: Very nice problem. This should lead you to read/learn about conics in terms of projective geometry (e.g., section 2 of chapter 8 of your algebra book :) ) It turns out conics are given by a projective (linear fractional) transformation from the pencil of lines through one point to the pencil of lines through another. The question is to characterize which type of conic you get by invariants of that projective transformation. $\endgroup$ – Ted Shifrin May 27 '14 at 21:33
  • 3
    $\begingroup$ See also gdaymath.com/lessons/quadratics1/6-2-light-cones $\endgroup$ – Rahul May 27 '14 at 21:49
  • 1
    $\begingroup$ @Rahul, that article is quite illuminating (pun intended). Thanks for the share! $\endgroup$ – Kaj Hansen May 27 '14 at 21:51
  • 3
    $\begingroup$ This was my homework... I will post my proof. $\endgroup$ – evil999man May 28 '14 at 4:22
  • 1
    $\begingroup$ math.stackexchange.com/questions/358056/… BTW I was able to explain this to my son when he was five. Unfortunately, he tried to tell his kindergarten teacher and was disappointed that she had no clue. $\endgroup$ – Ron Gordon May 28 '14 at 21:31

Oh... this question was my homework a year ago, and this is my proof which is short as per my laziness.

enter image description here

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.

  • 9
    $\begingroup$ If you take the definition of hyperbola as intersection of cone and plane as two parts are made, this is fairly obvious. $\endgroup$ – evil999man May 28 '14 at 4:42

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!

  • 1
    $\begingroup$ Both a great question, and a great answer! +1 to both, enjoyed reading them. $\endgroup$ – recursive recursion May 27 '14 at 21:21
  • 3
    $\begingroup$ Nices answer, but you could have guessed: the conic projection of a circle is a conic, and here it's obviously unbounded. $\endgroup$ – Jean-Claude Arbaut May 27 '14 at 21:28
  • 3
    $\begingroup$ The intersection of a right cone with a plane not through the cone's vertex is a circle if the plane is perpendicular to the axis of the cone, and a parabola if the plane is parallel to a plane through the vertex, tangent to the cone. Any other plane that produces a bounded conic produces an ellipse, and all other planes produce hyperbolas. But observation of the shadow a lampshade casts on a wall is a great way to motivate the study of conics. $\endgroup$ – David K May 27 '14 at 21:44
  • 2
    $\begingroup$ In fact, isn't that what the mathematical term "obviously" means? :-) $\endgroup$ – David K May 27 '14 at 21:47
  • 1
    $\begingroup$ Greatly enjoyed the feedback everyone. Related to David's comment, an approach similar to mine could probably be used to prove that each of the different ways of slicing a cone will yield its corresponding conic section. Of course, if you already have those results, then Jean-Claude's approach is a lot faster. $\endgroup$ – Kaj Hansen May 27 '14 at 23:05

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.

  • $\begingroup$ After some searching, it does appear that this is a pretty common example. It was a cool experience for me though since I'd never seen it before. $\endgroup$ – Kaj Hansen May 28 '14 at 8:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.