Find the standard form of the equation of the parabola with a focus at (0, 4) and a directrix at y = -4.

Question

imstuck · Answer

First you need to decide if this is an x^2 parabola or a y^2 parabola, and then if it is a positive or a negative type.

wade123 · Answer

it is y^2

wade123 · Answer

and it would be negative right?

imstuck · Answer

No it's not a y^2.  The point plotted and the directrix looks like this on a graph:

imstuck · Answer

|dw:1408123780524:dw|

wade123 · Answer

ohh

imstuck · Answer

The graph ALWAYS opens up around the focus.  In other words, it encloses the focus, so this one opens upwards around the y axis.  So it is an x^2 parabola, and a positive one at that.  Positive ones open upwards like a cup.

wade123 · Answer

gotcha!!

imstuck · Answer

The rule for the vertex is that is it is halfway between the focus and the directrix.  Right here on your graph:

imstuck · Answer

|dw:1408124001872:dw|

wade123 · Answer

so for my answer i got y=1/16x^2

imstuck · Answer

So this is your vertex, at (0,0), the origin.  The formula for the parabola of this type is $$(x-h)^{2}=4p(y-k)$$

wade123 · Answer

oops it should be 4 not 16

imstuck · Answer

Ok, then let me check your answer...give me just one sec, ok?

wade123 · Answer

okk