Question

Find the vertex, focus, directrix, and focal width of the parabola
-1/16 x^2=y

Answer 1

Were you having trouble with one of these in particular? Or not sure how to begin with it?

Answer 2

For the vertex, we can quickly see that this is in vertex form:
(y - k) = a (x - h)^2
With (h, k) at (0, 0).

For the focus and directrix, we could use the relationship between a and the distance between the focus-vertex and directrix-vertex. |a| = 1/4|p|.
The focus lies on the inside of the cave created by the parabola. The directrix lies behind it. Both distances are calculated along the axis of symmetry for the parabola, which is a vertical line.
To get to directrix (which is above because the parabola opens downward), we find the horizontal (perpendicular to axis of symm) line p units above the vertex. The focus is p units below the vertex.

For focal width, I believe this is just 4*p.

Answer 3

If you still need help, please bump your question up or tag some helpers! I will be away! Good luck with the Maths. :)

Answer 4

i i tried and i got

Answer 5

focus=-8,0

Answer 6

The vertex of x^2 = -16y is (0, 0). The coefficient of the unsquared part is -16, and this is also the value of 4p, so p = –4. Since the x part is squared and p is negative, then this is a regular parabola that opens downward. This means that the directrix, being on the outside of the parabola, is four units above the vertex, and the focus is four units below the vertex. The focal width, aka the focal parameter, is in general 4*|p|.
Vertex (0, 0); Focus (0, -4); Directrix y = 4, Focal width = 8

Answer 7

dirextes=4

Answer 8

i got 64 as the focal width

Answer 9

How did you get that? Look at my rewrite of your original equation. Is that correct?

Answer 10

Is this your original?\[-\frac{ x^2 }{ 16 }=y\]