anonymous
  • anonymous
If a particular radio telescope is 100ft in diameter and has a cross section modeled by the equation x^2=167y, how deep is the parabolic dish? What is the location of the focus? can someone help me with this?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
learned a bit more about this stuff today i did....
amistre64
  • amistre64
x^2 = 4ay where a is the distance from the vertex..... and your vertex is at 0,0
amistre64
  • amistre64
167/4 is a so the focus is at..... 0+ 167/4

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More answers

anonymous
  • anonymous
right. i have that its the first part that im having trouble with
amistre64
  • amistre64
ok.....we know that it 100 feet across, so that is 50 feet in both directions, when y = 50, what is x^2 = ?
amistre64
  • amistre64
is it headed in the x direction or the y?
anonymous
  • anonymous
the parabola opens up
amistre64
  • amistre64
then I would turn it into the normal way of viewing this: y =(1/167)x^2. when x = 50, whats y=?
amistre64
  • amistre64
2500/167...
anonymous
  • anonymous
so thats how deep it is?
amistre64
  • amistre64
yup,...
amistre64
  • amistre64
the numbers are the same, we just repositioned them into the "normal" way that this equation would look.
amistre64
  • amistre64
the vertex is at (0,0) thats half way, half of the diameter is 50...plug it in :)
amistre64
  • amistre64
14.97 is what I get
anonymous
  • anonymous
k I'll try that. thanks!
amistre64
  • amistre64
youre welcome :)
amistre64
  • amistre64
have you ever worked with quadratic equations?
amistre64
  • amistre64
cause thats all a parabola is...
anonymous
  • anonymous
yes. im in pre calc at a jr college, this stuff is being thrown at me quick but im doing my best.
amistre64
  • amistre64
phcc sound familiar to you?
anonymous
  • anonymous
no whats that?
amistre64
  • amistre64
its the community college I go to right now :) just wondering if you were in my trig/pre calc class ....
anonymous
  • anonymous
na im at tcc
amistre64
  • amistre64
well, good luck to ya :)
anonymous
  • anonymous
thanks!

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