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Elliptical Equation of a Spotlight

  1. Jun 22, 2009 #1
    1. The problem statement, all variables and given/known data
    I've pretty much given up on this problem:

    "A spotlight throws a beam of light that is 25 cm in diameter. If the beam hits the stage floor at an angle of 60 degrees with the horizontal, find an equation for the elliptical pool of light on the stage floor. (Figure 2-23)"

    http://books.google.com/books?id=4lwuyKlK6FsC&lpg=PA50&ots=cJyn8pzoNu&dq=33&pg=PA50" at Google books.

    The answer is [tex](x^2/208.3) + (y^2/156.3) = 1 [/tex] (all odd-numbered questions have answers provided in the back). It doesn't show any work, however.

    2. Relevant equations

    The equation for an ellipse:

    [tex](x^2/a^2) + (y^2/b^2) = 1[/tex]

    Equation for circle:

    [tex]x^2 + y^2 = a^2[/tex]

    Pythagorean Theorem

    [tex]a^2 = b^2 + c^2[/tex]

    3. The attempt at a solution

    I tried

    [tex]25*cos (60) = 12.5 [/tex]

    to attempt to find the length of the y axis above the horizontal. This is correct according to the back of the book as

    [tex]12.5^2 = 156.25[/tex].

    But now I'm stuck.
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Jun 22, 2009 #2

    Dick

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    You got the wrong number right by accident. The a and b in your equation are like 'radii' not 'diameters'. 156.25=(25/2)^2=(12.5)^2 which is the axis that doesn't get multiplied by any trig function. Because it doesn't change. What's the longer one?
     
  4. Jun 23, 2009 #3
    I agree that the attempted solution should be incorrect because of what you described, but it got the "magic number" (which could have been deduced more properly by theorizing that one axis stays the same length as the diameter of the beam of light). This brings me back to the same question ; it seems (to me at least) there isn't enough information to find the length of the larger axis.
     
    Last edited: Jun 23, 2009
  5. Jun 23, 2009 #4

    Dick

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    Sure there is. At the point where the 'back' of the beam hits the floor (the one closest to the light) draw a diameter across the beam. That's perpendicular to the beam so it makes one leg of a right triangle with length 25 cm with the length you are looking for being the hypotenuse. Find an angle and use trig.
     
  6. Jun 23, 2009 #5
    [tex]1/cos (30) = x/25[/tex]

    [tex] x/2 = 14.43[/tex]

    Substitute for a in the equation

    [tex](x^2/208.3) + (y^2/156.3) = 1 [/tex]

    which is the answer. Thanks :biggrin:.
     
    Last edited: Jun 23, 2009
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