Question

i need help from @Luigi0210 and @halorazer

Answer 1

yay c: can you help me with some questions

Answer 2

Find the vertex, focus, directrix, and focal width of the parabola.

negative 1 divided by 40 x squared equals y

Answer 3

okay soooooothe vertex would be 0,0 right?

Answer 4

nopes

Answer 5

how did u calculate it ?

Answer 6

i put it into a calculator .-. the parabola runs straight through the point 0,0..... plus all of my choices state that 0,0 is the vertex cuz its the only choice

Answer 7

oh, vertex, yes thats 0,0
i was looking for focus :P

Answer 8

XD okay the focus is 0,-10

Answer 9

correct :)

Answer 10

the diretrix is 10

Answer 11

and the focal length is 40 c:

Answer 12

wait just kidding focal length is 160...

Answer 13

directrix is "y=10" correct :)
i forgot how to calculate focal length :P

Answer 14

oh .-. hummmm let me look it up on google
\

Answer 15

lol, then how did YOU calculate it ?? :O

Answer 16

its more of a, eye the graph and guess kinda thing c:

Answer 17

focal length is distance between focal point to vertex!

Answer 18

so it would be 0-(-10) = 10 :)

Answer 19

well the only answer with a focal length of 10 has a diretrix of -10 and focus of 0,10

Answer 20

focus = 0,-10
directrix : y=10
are surely correct, so choose that option

Answer 21

but the focus on the question say +10 and you keep putting -10 and they keep say the diretrix is -10 and you keep saying its +10 .-. youre like completely opposite

Answer 22

Vertex: (0, 0); Focus: (0, -10); Directrix: y = 10; Focal width: 160

Vertex: (0, 0); Focus: (-20, 0); Directrix: x = 10; Focal width: 160

Vertex: (0, 0); Focus: (0, -10); Directrix: y = 10; Focal width: 40

Vertex: (0, 0); Focus: (0, 10); Directrix: y = -10; Focal width: 10

Answer 23

those are my choices .-.

Answer 24

who are 'they' ??
let me confirm online...
http://www.wolframalpha.com/input/?i=y%3D+%28-1%2F40%29x%5E2
click on "properties"

Answer 25

the answer people :P

Answer 26

so then if that says that then the focal length obviously has to be eitehr 40 or 160

Answer 27

for \(y = 4ax^2\)
focal width = 4a

Answer 28

the question is \(y =(-1/40)x^2\)
right ?

Answer 29

yes c:

Answer 30

a =-40
so focal width = |4*-40| =|- 160| = 160
:)

Answer 31

yay c: thank you ^.^

Answer 32

welcome ^_^