SOLUTION: A cross section of the mirror of a flashlight is a parabola. It measures 6 cm across and has a depth of 2 cm. How far from the vertex should the filament of the light bulb be pla

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A cross section of the mirror of a flashlight is a parabola. It measures 6 cm across and has a depth of 2 cm. How far from the vertex should the filament of the light bulb be pla      Log On


   



Question 1206394: A cross section of the mirror of a flashlight is a parabola. It measures 6 cm across and has a depth of 2 cm. How far from the vertex should the filament of the light bulb be placed for the flashlight to have its beam run parallel to the axis of its mirror?
A. 2.667 cm
B. 0.889 cm
C. 1.125 cm
D. 2.25 cm
E. 0.444 cm
F. 0.75 cm

Found 2 solutions by ikleyn, MathLover1:
Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.
A cross section of the mirror of a flashlight is a parabola. It measures 6 cm across
and has a depth of 2 cm. How far from the vertex should the filament of the light bulb
be placed for the flashlight to have its beam run parallel to the axis of its mirror?
A. 2.667 cm
B. 0.889 cm
C. 1.125 cm
D. 2.25 cm
E. 0.444 cm
F. 0.75 cm
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Place the origin of the coordinate system at vertex of the parabola.
Use the vertex form equation of this parabola 

    y = ax%5E2.        (1)


At x = 3 cm, we are given y = 2 cm.  So, this vertex equation takes the form

    2 = a*3^2,  which gives  a = 2%2F9.


So, equation (1) is  

    y = %282%2F9%29%2Ax%5E2.    (2)


For the parabola with equation y = ax%5E2,  the distance from the vertex to the focus is

    f = 1%2F4a.


In our case, the distance from the vertex to the focus is   1%2F%284%2A%282%2F9%29%29 = 9%2F8 cm = 1.125 cm.


ANSWER.  The distance from the vertex to the focus, where the filament should be placed, is 1.125 cm.  

         Option (C).

Solved.



Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

Its diameter is 6cm and its depth is 2cm
for simplicity, let vertex be at (0,0)
parabola opens to the right, and we use the standard equation in the form
y%5E2=4p%2Ax where focus is at p
two points on this parabola are (2,3) and (2,-3) => diameter is 6cm, 3cm above and 3cm below x-axis and depth is x=2cm
using one point we can calculate p
3%5E2=4p%2A2
9=8p
p=9%2F8
p=1.125


answer: C.