Find the standard form of the equation of the parabola with a focus at (0, 2) and a directrix at y = -2.

Question

goformit100 · Answer

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whpalmer4 · Answer

@jhannybean @goformit100 What does a directrix of y=-2 mean to you?

whpalmer4 · Answer

@jhannybean @goformit100 Perhaps a sketch showing both your proposed parabola and y = -2 would be informative :-)

whpalmer4 · Answer

Just draw me a picture, okay? :-)

whpalmer4 · Answer

Draw me a picture showing y = -2 as the directrix with the parabola you specified...

whpalmer4 · Answer

Okay, I'll plot it.

whpalmer4 · Answer

I mislabeled the parabola — though it says $y^2=x$ it is a graph of $y^2=8x$

jhannybean · Answer

What do you use to show your graphs btw?

whpalmer4 · Answer

Mathematica

jhannybean · Answer

Ahh, ok.

whpalmer4 · Answer

Does $y=-2$ look like the directrix of that parabola? :-)

jhannybean · Answer

Oh i see...it would have been x=-2.... Hmm.

jhannybean · Answer

I'll delete my work. lol.

whpalmer4 · Answer

Why not post a corrected version?

jhannybean · Answer

http://www.sketchtoy.com/39052824 he distance from the focus to the vertex is equal to the distance from the directrix to the vertex. The distance is labeled as "p" Use the form $$\large (x-h)^2 = 4p(y-k)$$ since this parabola is opening up and down. vertex= (0,0) ,p = 2 (distance of center from focus and center to directrix) $$\large (x-0)^2 = 4(2)(y-0)$$$$\large x^2 = 8y$$ And your final graph will look something like http://www.sketchtoy.com/39073173