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

-(1/16)x^2 = y

Question

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

anonymous · Answer

OMG I was just working on this with my friend!!!
Okay, so do you have any ideas how to solve this problem?

anonymous · Answer

Not really.

anonymous · Answer

Okay.... Well, do you know any definitions?

anonymous · Answer

I think I know the right answer. I just have an odd way of solving it, and I wanted someone to verify it. I got; vertex (0,0); focus (0,-4) directrix y=4; focal width=16

anonymous · Answer

Yes! That is correct! How do you solve it?

anonymous · Answer

I have an odd manner and it's complicated to explain, most of it I do in my head.

anonymous · Answer

Thanks for the help

anonymous · Answer

Ah. Okay. I'm all for weirdo methods! Always do those kinds of things at my school! I solve it by plotting the vertex, moving down the P (see formula), and moving up the p for the directrix, and then if the parabola is up and down, the line is horizontal, and vice versa!

No problem! I LOVE this topic, so let me know if you have anymore questions with conics!