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anonymous

  • one year ago

Pre-calc help? Find the vertex, focus, directrix, and focal width of the parabola. x^2 = 20y A. Vertex: (0, 0); Focus: (0, 5); Directrix: y = -5; Focal width: 20 B. Vertex: (0, 0); Focus: (5, 0); Directrix: x = 5; Focal width: 5 C. Vertex: (0, 0); Focus: (5, 0); Directrix: y = 5; Focal width: 80 D. Vertex: (0, 0); Focus: (0, -5); Directrix: x = -5; Focal width: 80

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  1. anonymous
    • one year ago
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    The equation of the parabola with vertex (h,k) is \[\Large (x-h)^{2}=4p(y-k)\] So is easy to see that the vertex is (0,0), So, the new equation is \[\Large x ^{2}=4py\] if 4p=20 then p=5 and the focus of a parabola is (h,k+p), in this case (0,0+5) = (0,5) and the directrix of a parabola is y=k-p, in this case 0-5 = -5 So the vertex is (0,0), the directrix is y=-5 and the focus is (0,5) Correct alternative is A

  2. anonymous
    • one year ago
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    Oh, I was just overcomplicating things by trying to put the equation into y^2 = 4px form!

  3. anonymous
    • one year ago
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    PS: The focal width is 4p, in this case 4x5=20

  4. Jhannybean
    • one year ago
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    note: if you cannot find a way to `complete the square`, then you will most likely not have a vertex.

  5. anonymous
    • one year ago
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    Yeah, sometimes it can be x^2 or y^2

  6. anonymous
    • one year ago
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    If no vertex exists, what do I put for the answer if a question asks for a vertex?

  7. anonymous
    • one year ago
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    Undefined?

  8. Jhannybean
    • one year ago
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    The vertex will exist, but it will most likely be defined at \((0,0)\) since the formula you're working off of is \((x-\color{red}{h})^{2}=4p(y-\color{red}{k})\) vertex = \((\color{red}{h,k})\)

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