Results 1 to 4 of 4

Math Help - Lamp with a parabolic reflector

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    8

    Arrow Lamp with a parabolic reflector

    I attempted to solve this problem for hours.
    I used x^2=4py and substitution. Didn't quite work.
    Please, help. Thanks!

    A lamp with a parabolic reflector is shown in the figure. The bulb is placed at the focus and the focal diameter is 20 cm. Find the diameter of the opening 15 cm from the vertex. Round to the nearest centimeter, if necessary.
    35
    38
    40
    39
    37
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2009
    Posts
    806
    Thanks
    4
    Quote Originally Posted by jeunesse27 View Post
    I attempted to solve this problem for hours.
    I used x^2=4py and substitution. Didn't quite work.
    Please, help. Thanks!

    A lamp with a parabolic reflector is shown in the figure. The bulb is placed at the focus and the focal diameter is 20 cm. Find the diameter of the opening 15 cm from the vertex. Round to the nearest centimeter, if necessary.
    Focal diameter = 4p = 20 cm. So p = 5 cm
    If F is the focus, FD = DM where M is the point on the diretrix. DM = 15 + 5 = 20. So FD = 20 cm.
    Distance between the focus and center of opening circle is 15 - 5 = 10 cm.
    So the radius R = 20^2 - 10^2 = 17.32. Hence
    d = 2*17.32 = 34.63 ~ 35 cm.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,909
    Thanks
    771
    Hello, jeunesse27!

    Who wrote this problem?
    The term "diameter" is usually reserved for circles.


    A lamp with a parabolic reflector is shown in the figure. . Right!
    The bulb is placed at the focus and the focal diameter is 20 cm. .
    latus rectum?
    Find the width CD of the opening 15 cm from the vertex, to the nearest cm.

    . . (a)\;35 \qquad (b)\;38 \qquad (c)\;40 \qquad (d)\;39\qquad (e)\;37
    I'll take a guess at what's going on . . .
    Code:
                    |
                  15|
        C o - - - - + - - - - o D
                    |
                   F|
    (-10,p)o- - - - o - - - -o (10,p)
                    |
            *       |p      *
             *      |      *
               *    |    *
          - - - - -*o*- - - - - - - - - -
                    V

    The equation is: . x^2 \:=\:4py
    . . where p is the distance from the vertex V to the focus F.

    The length of the latus rectum is 4p.
    . . Hence: . p \,=\,5

    The equation is: . x^2 \:=\:20y


    If y = 15, we have: . x^2 \:=\:20(15) \:=\:300 \quad\Rightarrow\quad x \:=\:\pm\sqrt{300}\:=\:\pm10\sqrt{3}


    Therefore: . CD \;=\;2(10\sqrt{3}) \;=\;34.64191615 \;\approx\;35\text{ cm}\;\;\;\text{ answer }(a)

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2009
    Posts
    8

    Thank you!

    Quote Originally Posted by sa-ri-ga-ma View Post
    Focal diameter = 4p = 20 cm. So p = 5 cm
    If F is the focus, FD = DM where M is the point on the diretrix. DM = 15 + 5 = 20. So FD = 20 cm.
    Distance between the focus and center of opening circle is 15 - 5 = 10 cm.
    So the radius R = 20^2 - 10^2 = 17.32. Hence
    d = 2*17.32 = 34.63 ~ 35 cm.
    Thanks to all of my helpers! The detailed explanation cleared the topic up for me!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Householder reflector
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 27th 2011, 03:23 AM
  2. Fringe pattern of a sodium lamp
    Posted in the Math Topics Forum
    Replies: 6
    Last Post: January 17th 2011, 12:11 AM
  3. parabolic tv antenna ( HELP!!! )
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 18th 2010, 05:46 PM
  4. parabolic
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: June 9th 2008, 07:36 AM
  5. parabolic arch
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: January 7th 2007, 05:52 AM

Search Tags


/mathhelpforum @mathhelpforum