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

Question

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

anonymous · Answer

lol everyone hates these conic section questions don't they ?
start maybe with 
$$x^2=-16y$$ which looks a lot like 
$$(x-h)^2=4p(y=k)$$ with $h=k=0$ meaning the vertex is at the origin $(0,0)$ and $p=-4$

anonymous · Answer

actually i had a typo there, should have been $(x-h)^2=4p(y-k)$

anonymous · Answer

i have this so far:
but i think im wrong, Vertex: (0, 0); Focus: (-8, 0)

anonymous · Answer

think the focus is wrong
$p=-4$ so go 4 units down from $(0,0)$ to get the focus

anonymous · Answer

Focus: (0, -4)?

anonymous · Answer

yup

anonymous · Answer

and directix is the horizontal line 4 units up

anonymous · Answer

Vertex: (0, 0); Focus: (0, -4); Directrix: y = 4; Focal width: 16

anonymous · Answer

could be, i have no idea what "focal width" is, but i could look it up

anonymous · Answer

hahha is eveyrhting else i did right?

anonymous · Answer

yes, and focal width is $|4p|=16$ so you are right there as well

anonymous · Answer

THANK YOU