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- 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.

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- anonymous

- katieb

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- 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
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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

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

- anonymous

oh is that the answer?

- 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

so its 4?

- 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

I am sorry, so I need the difference between the focus and the vertex, is it 4?

- 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

awesome! thanks

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