Question 636443
Give the vertex, focus, and directrix of the parabola 
16x^2=y
**
Form of equation for a parabola that opens upwards: x^2=4py
For given problem:
x^2=(1/16)y
4p=1/16
p=1/64
Equation: 
x^2=y/64
..
vertex: (0,0)
axis of symmetry: x=0
focus: (0, 1/64) (p distance above vertex on the axis of symmetry)
directrix: y=-1/64 (p distance below vertex on the axis of symmetry)