Question

Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, -4).

Answer 1

y=(-1/4)x^2 if it opens vertically

Answer 2

x=(-1/4)y^2 if it opens vertically

Answer 3

Not quite right :P

Answer 4

\[y=-\frac{ 1 }{ 4 }x ^{2}\]

Answer 5

I know she meant horizontally for her second post, but It's going to open vertically for certain. Do you care for an explanation, or just kinda want the answer?

Oh, and the x^2 and y^2 term in the standard form equation has to be positive, you can't have it being multiplied or positive or anything like that.

Answer 6

*or negative

Answer 7

thats what i got. and i would like an explanation so i dont have to ask for help in the future

Answer 8

So starting with the correction I was mentioning, you are going to either have y^2 = 4px or x^2 = 4py. The x^2 and y^2term is always by itself, not being multiplied, not negative, none of that. Of course x^2 means it may open up or down, y^2 means it may open left or right. The way you tell is using the focus and the directrix. So now for the info that's important.

The directrix is always at -p and the focus is always at p.

So this problem says the vertex is at the origin and the focus is at (0,-4). This means that p is -4. Also know that since the focus is always inside of the parabola, this parabola will open down. Since we only need the equation, we don't necessarily need the directrix or anything like that.

Now recall, the formula is x^2 = 4py. Well, if p is -4, then 4p must be -16. So actually, the answer would be x^2 = -16y.

Answer 9

Forgive the horrible drawing, but thought I'd post that. Because as you can see, you can also use the value of p to see how wide the parabola is. The entire width is equal to 4p, 2p units left and 2p units right. Of course this would be 2 units up and down if it were a y^2 problem, but yeah xD