SOLUTION: What is the vertex, directrix, and focal width of the parabola: x^2=36y

Question 585360: What is the vertex, directrix, and focal width of the parabola: x^2=36y
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
What is the vertex, directrix, and focal width of the parabola: x^2=36y
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Standard form of equation for a parabola that opens upwards: (x-h)^2=4p(y-k), (h,k) being the (x,y) coordinates of the vertex.
For given equation: x^2=36y
vertex: (0,0)
axis of symmetry: y-axis or x=0
4p=36
p=9
directrix: y=-9 (p units below the vertex on the axis of symmetry)
focus: (0,9)
focal width=4p=36
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