SOLUTION: a flashlight is shaped like a paraboloid so that if its bulb is places at then focus, the light rays from the bulb will then bounce off the surface in a focused direction that is p

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: a flashlight is shaped like a paraboloid so that if its bulb is places at then focus, the light rays from the bulb will then bounce off the surface in a focused direction that is p      Log On


   

Question 1167082: a flashlight is shaped like a paraboloid so that if its bulb is places at then focus, the light rays from the bulb will then bounce off the surface in a focused direction that is parallel to the x-axis. if the paraboloid has a depth of 1.8 inches and the diameter on its surface is 6 inches, how far should the light source be placed from the vertex
Answer by ikleyn(53937) About Me  (Show Source):
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a flashlight is shaped like a paraboloid so that if its bulb is places at then focus,
the light rays from the bulb will then bounce off the surface in a focused direction
that is parallel to the x-axis. if the paraboloid has a depth of 1.8 inches
and the diameter on its surface is 6 inches, how far should the light source be placed from the vertex
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        It is a standard problem of this kind. The feature is
        that in this problem the symmetry line is parallel to x-axis,
        while usually in such problems the symmetry line is parallel to y-axis.
        So, I will adapt a standard solution to this case. 


For solving such problems, write an equation of the parabola in the cross-section
in the form

    x = %281%2F%284p%29%29%2Ay%5E2.    (1)


The advantage of writing in this form is the fact that then "p"
is the distance from the parabola vertex to its focus.


So, now our task is to find value of "p" from the given data.


The fact that "the paraboloid has a depth of 1.8 inches 
and the diameter on its surface is 6 inches"
means that x = 1.8 meters at y = 6/2 = 3 inches.


So, we substitute this data into equation (1), and we get

    1.8 = %281%2F%284p%29%29%2A3%5E2.


It gives

    p = %281%2F4%29%2A%28%283%5E2%29%2F1.8%29 = 1.25 = 11%2F4 inches.


Thus, the light source should be placed 11%2F4 inches from the vertex.    ANSWER

Solved.