anonymous
  • anonymous
A radio telescope has a parabolic surface as shown below. A parabola opening up with vertex at the origin is graphed on the coordinate plane. The height of the parabola from top to bottom is 1 meter and its width from left to right is 8 meters. If the telescope is 1 m deep and 8 m wide, how far is the focus from the vertex? I will post a pic in a second.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
anonymous
  • anonymous
campbell_st
  • campbell_st
well a basic form of the parabola is \[(x - h)^2 = 4a(y - k)\] (h, k) is the vertex and a is the focal length |dw:1438028123648:dw| so the vertex is at the origin so the equation is \[x^2 = 4ay\] substitute a point from the curve (4, 1) into the equation to find the focal length \[4^2 = 4 \times 1 \times a\] ad the the focus is a units about the vertex hope it helps

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anonymous
  • anonymous
alright I needed some help, not for someone to write a book lol. 16 = 4 * 1 * a 16 = 4a a = 4 Now what
anonymous
  • anonymous
oh is that the answer?
campbell_st
  • campbell_st
that is the focal length, the distance from the vertex to the focus.... so if the vertex is (0, 0) the focus is at (0, 0+a)
anonymous
  • anonymous
so its 4?
campbell_st
  • campbell_st
well the y value in the focus is y = 4 the x value is the line of symmetry x = 0 so the focus is at (0, 4)
anonymous
  • anonymous
I am sorry, so I need the difference between the focus and the vertex, is it 4?
campbell_st
  • campbell_st
opp... sorrry you are right... I misread the question as where is the focus... the focal length is 4 so the focus is 4 metres above the vertex
anonymous
  • anonymous
awesome! thanks

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