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anonymous
 one year ago
Find the vertex, focus, directrix, and focal width of the parabola.
(1/16)x^2 = y
anonymous
 one year ago
Find the vertex, focus, directrix, and focal width of the parabola. (1/16)x^2 = y

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OMG I was just working on this with my friend!!! Okay, so do you have any ideas how to solve this problem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay.... Well, do you know any definitions?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think I know the right answer. I just have an odd way of solving it, and I wanted someone to verify it. I got; vertex (0,0); focus (0,4) directrix y=4; focal width=16

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes! That is correct! How do you solve it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have an odd manner and it's complicated to explain, most of it I do in my head.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah. Okay. I'm all for weirdo methods! Always do those kinds of things at my school! I solve it by plotting the vertex, moving down the P (see formula), and moving up the p for the directrix, and then if the parabola is up and down, the line is horizontal, and vice versa! No problem! I LOVE this topic, so let me know if you have anymore questions with conics!
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