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

Question

campbell_st · Answer

the focus is below the vertex so the focal length is -4

a form of the parabola is 

$$x^2 = 4ay$$

you know a = -4 so substitute and make y the subject

anonymous · Answer

y=-1/4x^2?

campbell_st · Answer

well not quite substitute a = -4 gives

$$x^2 = -16 y$$

now make y the subject

jdoe0001 · Answer

notice one thing, the vertex is at the origin, that is, (0 , 0)
the focus is at (0, -4), lemme graph that |dw:1381614452158:dw|

so as you can see, from the vertex to the focus, there's a distance of "4 units"
notice the parabola is opening DOWNWARDS, that means a negative focus unit, or -4

and is also moving over the y-axis

the equation for a parabola moving over the y-axis using the "focus form" of it will be $\bf (x-h)^2=4p(y-k)$

(h, k)  = vertex coordinates
   p      = distance from the vertex to the focus

so plug the values in, and solve for "y" to get the equation

anonymous · Answer

oh my bad y=-1/16x^2

campbell_st · Answer

thats it