SOLUTION: I have to write the standard form of the equation of the parabola with its vertex at (0,0) and directrix y=-3/2. I've tried looking in my algebra 2 book, should be within chapter

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: I have to write the standard form of the equation of the parabola with its vertex at (0,0) and directrix y=-3/2. I've tried looking in my algebra 2 book, should be within chapter       Log On


   



Question 864826: I have to write the standard form of the equation of the parabola with its vertex at (0,0) and directrix y=-3/2.
I've tried looking in my algebra 2 book, should be within chapter ten and I don't see any examples showing this type of question. I've tried looking up examples online and all I can find is writing the standard form equation with two of the parabolas points or with the focus and directrix, never the vertex and directrix. Is this the same as focus and directrix? Please explain.

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
write the standard form of the equation of the parabola with its vertex at (0,0) and directrix y=-3/2.
given data shows that parabola opens up. (directrix is below the vertex of the parabola)
Its basic equation: (x-h)^2=4p(y-k), (h,k)=coordinates of vertex
since vertex is at the origin, basic equation is x^2=4py
For given parabola, p=3/2 (distance from vertex to directrix (and focus) on the axis of symmetry)
4p=6
Standard form of the equation for given parabola: x^2=6y