1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Parabolas math problem

  1. Mar 29, 2005 #1
    I'm having difficulty with this question. All help is appreciated.

    *The cross section of television antenna dish is a parabola and the receiver is located at the focus.

    A. If the receiver is located 5 feet above the vertex, assume the vertex is the origin, find an equation for the cross section of the dish.
    Okay, I know the vertex is 0,0. The focus is 0, 5. The equation is x^2=4ay.
    I don't know where to go from there, or what equation is needed to find the cross section.
     
  2. jcsd
  3. Mar 29, 2005 #2
    Actually, I figured it out. x^2 = 4ay, and a must equal 5 because the focus is (0,5).
    That means teh equation is x^2 = 4(5)y or x^2 = 20y.

    What I can't figure out is part B:
    If the dish is 10 feet wide, how deep is it?
    I have never had a question like this before. How do you know how "deep" a dish is?
     
  4. Mar 29, 2005 #3
    So the equation of the parabola is [itex]y=x^2/20[/itex]. If it's 10 feet wide and centered at the origin, then it's cross section is between -5 and 5 on the x-axis. So, to find the depth, you need to calculate "y" for x=5... that is, if I understand the question correctly.

    - Kamataat
     
  5. Mar 29, 2005 #4
    Thank you for the help. I think that's right. Solving for y, it would be 1.25, which is the answer. I just didn't know how to come to it and show my work. Thanks again.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Parabolas math problem
  1. Is this parabola? (Replies: 7)

  2. Math problem (Replies: 3)

Loading...