Write the equation in standard form for the parabola with vertex (0, 0), and focus (0, 1).

Question

anonymous · Answer

I got y=1/2 which isn't even on the answer option list

anonymous · Answer

It's asking for the equation. o.o

anonymous · Answer

The answer options are:
1.)x=1/2y^2
2).4y^2
3)y=4x^2
4)y=1/2x^2

anonymous · Answer

I know it is asking for the equation

kohai · Answer

$$(x-h)^2 = 4p(y-k)$$

(h, k) as the vertex
p = distance from the vertex to the focus
    p > 0, opens upward
    p < 0, opens downward

anonymous · Answer

What?

anonymous · Answer

They gave you the equation. Now all you have to do is plug the numbers you gave us into it.

anonymous · Answer

Hhaha ok that's not the equation I was taught

anonymous · Answer

And I don't know how to plug it in to that I've never seen it before

kohai · Answer

What equation were you taught?

anonymous · Answer

the distance formula

kohai · Answer

I'm not sure how you would use the distance formula for this, because you would just get d = some value

kohai · Answer

And all of the sites I'm looking at for this kind of question uses the formula I gave

anonymous · Answer

I don't know I'm really confused

anonymous · Answer

You don't use the distance formula to get an equation for a parabola. O.o

anonymous · Answer

Sorry but my textbook does

anonymous · Answer

kohai · Answer

Can you do me a favor and take a picture of the example your textbook does for this problem? So I can see what you're looking at?

anonymous · Answer

√(x-x1)^2+(y-y1)^2=√(x-x2)^2+(y-y^2)^2

kohai · Answer

Oh, I see.
$$\sqrt{(x-x1)^2+(y-y1)^2} = \sqrt{(x-x2)^2+(y-y2)^2}$$
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