Question

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

Answer 1

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

Answer 2

Oh, I was just overcomplicating things by trying to put the equation into y^2 = 4px form!

Answer 3

PS: The focal width is 4p, in this case 4x5=20

Answer 4

note: if you cannot find a way to `complete the square`, then you will most likely not have a vertex.

Answer 5

Yeah, sometimes it can be x^2 or y^2

Answer 6

If no vertex exists, what do I put for the answer if a question asks for a vertex?

Answer 7

Undefined?

Answer 8

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 = \((\color{red}{h,k})\)