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anonymous

  • one year ago

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.

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @Loser66

  3. campbell_st
    • one year ago
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    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

  4. anonymous
    • one year ago
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    alright I needed some help, not for someone to write a book lol. 16 = 4 * 1 * a 16 = 4a a = 4 Now what

  5. anonymous
    • one year ago
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    oh is that the answer?

  6. campbell_st
    • one year ago
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    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)

  7. anonymous
    • one year ago
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    so its 4?

  8. campbell_st
    • one year ago
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    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)

  9. anonymous
    • one year ago
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    I am sorry, so I need the difference between the focus and the vertex, is it 4?

  10. campbell_st
    • one year ago
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    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

  11. anonymous
    • one year ago
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    awesome! thanks

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