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I ran into this problem in a textbook and was intrigued by it. Conics are generally formed through different cuts one can make with the shape of a cone. But, there have been recent discussions on creating conical shapes through aiming a flashlight in a certain direction, which is interesting.

If you hold a flashlight parallel to the ground, the beam would be able to create a parabolic shape on the ground, which is one of the three conical shapes that can be made through a flashlight. Similarly, if one tilts a flashlight by aiming it on the wall, it could create an elliptical shape. However, I'm uncertain about what one can do to create a hyperbolic shape with a flashlight.

This picture is the reference for this problem and asks two things which I'm wondering if anyone has input on: (a) Why is the boundary of this lighted area a hyperbola? (b) How can one hold a flashlight so that its beam forms a hyperbola on the ground?

enter image description here

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    $\begingroup$ The picture at the end of the question doesn't even remotely resemble anything that you could see in reality (it is just some bad computer graphics). $\endgroup$
    – Marc van Leeuwen
    Mar 16, 2015 at 12:51

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This picture might be a good supplement to Neal's explanation, which is spot-on. If you imagine that your flashlight is pointing straight down from the tip of the cone, then the wall is any one of the colored areas, depending on the angle you hold it at.

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    $\begingroup$ Thanks, this is a good representation. One question though: what's the difference between the parabola and the hyperbola in the picture? $\endgroup$
    – Shrey
    Mar 16, 2015 at 3:57
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    $\begingroup$ No problem! Good question - the parabola is the exceptional borderline case. Look closely and you'll see that the parabola's axis is parallel to the cone's side. If it were even slightly angled further in either direction, it would become either a hyperbola or (part of) an ellipse, depending on which way it was tilted. As for what the qualitative differences are, the one that strikes me first is that ellipses are bounded and hyperbolae have asymptotes, but neither is true for parabolae. $\endgroup$
    – BGreen
    Mar 16, 2015 at 4:07
  • $\begingroup$ @ShreyDesai Circle is a section by a plane perpendicular to the cone's axis, parabola is a section by a plane parallel to the axis. Ellipse is anything between circle and parabola, hyperbola is anything between parabola and parabola (remember the cone has its mirror extension above the apex and similar set of sections there). And there are degenerate sections (a point, a ray, a line and a pair of lines) obtained when the cutting plane meets the apex. $\endgroup$
    – CiaPan
    Mar 16, 2015 at 10:40
  • $\begingroup$ @CiaPan As a commenter just pointed out to me in my answer above, a parabola is not a section parallel to the axis, it is a section parallel to the opposite side. ;) $\endgroup$
    – Neal
    Mar 16, 2015 at 11:43
  • $\begingroup$ @Neal That's right, parabola is a section obtained with a cutting plane parallel to the line contained in the conic surface (is it called a 'generatrix'?). A plane parallel to the cone axis gives a hyperbola as a section. $\endgroup$
    – CiaPan
    Mar 16, 2015 at 12:11
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Start with the flashlight parallel to the wall. If you incline it slightly away from the wall, the (conical) beam will still hit the wall and you'll get a hyperbola. It is only after the angle of the flashlight to the wall is greater than the angle of the light cone that the hyperbola vanishes at infinity. (In practice, of course, it is rather before then because the air will scatter the light.)

As CiaPan pointed out in the comments below, the criterion for being a parabola is that the cut be parallel to the opposite side of the light cone. The axis of the cone is the direction of the flashlight. Therefore you still get a hyperbola for the flashlight inclined toward the wall at shallow angles.

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    $\begingroup$ My comment automagically disappeared (possibly I accidentaly clicked the 'delete this comment' mark?), but Neal has already cited my main point, so I'm not going to re-invent what I wrote before. :) $\endgroup$
    – CiaPan
    Mar 17, 2015 at 10:04

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