Focus of parabola, checking my answer.

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jkristia
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I just had this question in an online test I took, and first I had 16ft, but for some unknown reason I chose to change it to 64ft right before I submitted the test, and of course I got it wrong. So I just want to confirm the answer is indeed 16ft.

Homework Statement



A parabolic antenna has a diameter of 32 feet and is 4 foot deep. How far is the focus from the vertex?

Homework Equations


The Attempt at a Solution



Handwritten-style worked solution: x^2 = 4ay, then 16^2 = 4a(4), then a = 16^2 / 16 = 16.


My answer - Focus for this parabola is (0, a), so (0, 16) or 16ft from vertex
 
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  • Handwritten-style worked solution: x^2 = 4ay, then 16^2 = 4a(4), then a = 16^2/16 = 16.
    parabola.png
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I think you have the depth and the diameter reversed.
 
Hmm, not sure I understand why you think that.

The way I read this is when x = 16, then y = 4, (the 'depth' 16 feet out to the bottom of the vertext is 4 feet), but maybe I misunderstood something.
 
jkristia said:
Hmm, not sure I understand why you think that.

The way I read this is when x = 16, then y = 4, (the 'depth' 16 feet out to the bottom of the vertext is 4 feet), but maybe I misunderstood something.

Never mind. I had my two variables interchanged. You are correct.