Question 136996
The equation of a parabola with vertex (h,k), axis of symmetry parallel to the y-axis, opening upwards, and distance from the vertex to the directrix of a is:


{{{(x-h)^2=4a(y-k)}}}


Rearranging your equation:


{{{y=(1/20) x^2}}}


{{{x^2=20y}}}


{{{(x-0)^2=4(5)(y-0)}}}


So you can see that {{{h = 0}}}, {{{k = 0}}}, and {{{a=5}}}


Therefore the vertex is (0,0).  The axis of symmetry, known to be parallel to the y-axis, passes through the vertex, therefore the axis of symmetry is the y-axis, or {{{x=0}}}.


The directrix is a line perpendicular to the axis of symmetry, a units distant from the vertex.  Since the axis of symmetry is a vertical line, the directrix is a horizontal line.  5 units below the vertex on the y-axis is the point (0,-5) and the directrix is the horizontal line intersecting this point, or {{{y=-5}}}.


The focus is a point on the axis of symmetry a units from the vertex, since the point is on the axis of symmetry, {{{x=0}}}, the x-coordinate of the focus must be 0.  Since the focus is {{{a=5}}} units from the vertex, the y-coordinate of the focus must be 5.  Therefore the focus is at (0,5)