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anonymous
 one year ago
Precalc 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
anonymous
 one year ago
Precalc 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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The equation of the parabola with vertex (h,k) is \[\Large (xh)^{2}=4p(yk)\] 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=kp, in this case 05 = 5 So the vertex is (0,0), the directrix is y=5 and the focus is (0,5) Correct alternative is A

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, I was just overcomplicating things by trying to put the equation into y^2 = 4px form!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0PS: The focal width is 4p, in this case 4x5=20

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0note: if you cannot find a way to `complete the square`, then you will most likely not have a vertex.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, sometimes it can be x^2 or y^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If no vertex exists, what do I put for the answer if a question asks for a vertex?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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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