SOLUTION: Please help me with this question! Find the vertex, focus, directrix, and focal width of the parabola. {{{ (-1/16) x^2 = y }}}

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Please help me with this question! Find the vertex, focus, directrix, and focal width of the parabola. {{{ (-1/16) x^2 = y }}}      Log On


   

Question 871289: Please help me with this question!
Find the vertex, focus, directrix, and focal width of the parabola.
+%28-1%2F16%29+x%5E2+=+y+
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Find the vertex, focus, directrix, and focal width of the parabola.
+%28-1%2F16%29+x%5E2+=+y+
x^2=-16y
This is an equation of a parabola that opens downward with vertex at the origin.
Its basic equation: x^2=-4py
For given parabola:
vertex: (0,0)
axis of symmetry: x=0
4p=16
p=4
focus: (0,-4)(p-distance below vertex on the axis of symmetry)
directrix:y=4 (p-distance above vertex on the axis of symmetry)
focal width=4p=16