Question 58915: Locating the receiver for a radio telescope.
The U.S. Naval research Laboratory designed a giant radio telescope weighting 3450 tons. Its parabolic dish has a diameter of 300 feet and a depth of 44 feet
(a) find an equation in the form y=ax^2 that describes a cross section of this dish.
(b) if the receiver is located at the focus, how far should it be from the dish(vertex)?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The U.S. Naval research Laboratory designed a giant radio telescope weighting 3450 tons. Its parabolic dish has a diameter of 300 feet and a depth of 44 feet
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(a) find an equation in the form y=ax^2 that describes a cross section of this dish.
Draw the parabolic cross section on a coordinate system.
Vertex at (0,-44); x-intercepts at (-150,0) and (150,0)
Form: (x-h)^2=4p(y-k)
h=0; k=-44; x=150; y=0
150^2=4p(44)
p=127.84
EQUATIOn:
x^2=4(127.84)(y+44)
y=(1/511.36)x^2 - 44
y=0.0019555695...x^2 -44
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(b) if the receiver is located at the focus, how far should it be from the dish(vertex)?
Focus at (0,p) = (0,127.84 ft.)
127.84 ft above the vertex
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Cheers,
Stan H.
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