SOLUTION: Can you help me find the vertex, focus, and equation of the directrix for x^2 =12y? From what I worked out I think the vertex is (0,0); the focus (0,3); and the equation is x^2=4*

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Can you help me find the vertex, focus, and equation of the directrix for x^2 =12y? From what I worked out I think the vertex is (0,0); the focus (0,3); and the equation is x^2=4*      Log On


   



Question 691661: Can you help me find the vertex, focus, and equation of the directrix for x^2 =12y?
From what I worked out I think the vertex is (0,0); the focus (0,3); and the equation is x^2=4*3y.

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
find the vertex, focus, and equation of the directrix for x^2 =12y?
This is an equation of a parabola that opens upwards
Its standard form: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of the vertex, p=distance from vertex to focus or directrix on the axis of symmetry
For given equation: x^2=12y
vertex: (0,0)
axis of symmetry:x=0 or y-axis
4p=12
p=3
focus: (0,3)
directrix: y=-3
you got everything right except the directrix which is just a horizontal straight line p-units below the vertex on the axis of symmetry