Parabolas...

S

shawie

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Aug 31, 2005
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wow... these parabolas are killing me. I know how to tell the vertex, focus, and directrix on a graph but for some reason i'm drawing a blank when it comes to giving the coordinates of a parabola's vertex and focus and the equation of its directrix from an equation.

For example:

y=-1/12x^2 how do i find the vertex? i know that the directrix is perpendicular to the focus but how do i find the focus?

help please? :(
 
You have to find the value for p and use that to figure the rest out, I can't remember it off hand but its the basic parabola formula, it has to be in your book. Use that formula, maybe someone else remembers it offhand.
 
A parabola parallel to the y-axis (vertical) with focal length \(\displaystyle \mbox{p}\) and vertex \(\displaystyle \mbox{(h, k)}\) has the equation
\(\displaystyle \mbox{ (x - h)^2 = \pm 4p(y - k) }\)

Shawie, can you get your equation into this form? (Note your parabola has h and k equal to 0).

The \(\displaystyle \mbox{\pm}\) is plus for upward-facing and minus for downward-facing.

Draw a sketch.
 
it means that the cooeficient of p determines if the parabola opens up or down
 
its always 4 becausae the "4" is part of the parabala formula. you use the given information to solve for the only unknown "p". I believe that p is the distance from the focus to the directrix but its been a while so I might be wrong, there has to be a picture in your book showing you what the variables in the parabola equation represent.
 
Actually I think p is the distance from the focus to the vertex, so 2p would be the distance from the focus to the directrix. But you better check on that.
 
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