Question

Find the vertex, focus, directrix, and focal width of the parabola. -1/20x^2=y PLEASE

Answer 1

@jdoe0001

Answer 2

so the parabola has an "x" SQUARED component, so we'd need to use the "focus form" for the parabola of \(\bf (x-h)^2=4p(y-k)\)

(h, k) = vertex point
p = distance from the vertex to the focus

Answer 3

so let's take a peek at yours \(\bf -\cfrac{1}{2}x^2 = y \implies -\cfrac{1}{2}(x-0)^2 = (y-0)\\
\textit{notice the (h, k) vertex}\\
\textit{notice the "p" distance } 4p = -\cfrac{1}{2}\)

Answer 4

hmm, wait a second... I got ... . a bit off there, lemme rewrite that

Answer 5

so p =-1/8?

Answer 6

well, from that yes, but the focus form isn't simplified

Answer 7

\(\bf -\cfrac{1}{2}x^2 = y \implies -\cfrac{1}{2}(x-0)^2 = (y-0)
\implies (x-0)^2 = -2(y-0)\\
\textit{notice the (h, k) vertex}\\
\textit{notice the "p" distance } 4p = -2\)

Answer 8

ahh okay so p is -1/2

Answer 9

yes, so the vertex is the origin
the focus is negative, means the parabola opens downward, with a hump

the focus is 1/2 from there over the y-axis DOWNWARDS, that is (0, -1/2)


directrix distance to the vertex is the same as the focus, 1/2, in the opposite direction

Answer 10

and the focus with is 4P

Answer 11

width rather

Answer 12

so how do i find what the focal width is??

Answer 13

is 4p

Answer 14

just bear in mind that "p" is neither negative nor positive, just a distance unit from the vertex

Answer 15

so the focal width will be 4p => 4( 1/2 )

Answer 16

I was thinking that but I got focal width as 2 and that wasn't on my options:/

Answer 17

that's why i'm a bit confused

Answer 18

what options do you have?