Question 194756
<font size = 8 color = "red"><b>Edwin's solution and graph:</b></font>
<pre><font size = 4 color = "indigo"><b>
I think you're right, except maybe you'd want to erase
the 0's in the final answer.

I'll go through it.

Draw the two points, vertex V(0,0) and the focus F(0,4)

{{{drawing(400,400,-10,10,-10,10, locate(.3,4.3,"F(0,4)"),

graph(400,400,-10,10,-10,10), line(-.2,4,.2,4),locate(.3,.8,"V(0,0)"),
locate(-.2,4.35,"@"),locate(-.2,.35,"@") )}}}

We can tell the parabola opens upward since the focus
is above the vertex.  The equation is 

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

Where the vertex is (h,k), and p is the directed
distance from the vertex to the focus.

If we put our pencil on the vertex and go to the focus
we move the pencil upward 4 units, so the parabolic constant
p = +4.  (When the pencil moves downward in going from the
vertex to the focus, then p is negative)

The directed distance from the vertex to the focus in
this problem is +4, and (h,k) = (0,0) so, substituting

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

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

{{{x^2 = 16y}}}

To draw the finished parabola, first draw the directrix
which is a horizontal line p units on the other side of the 
vertex.  In this case it is 4 units below the vertex.  It
has the eqution {{{y=-4}}}

{{{drawing(400,400,-10,10,-10,10, locate(.3,4.3,"F(0,4)"),
locate(2,-4,DIRECTRIX), 
graph(400,400,-10,10,-10,10), line(-.2,4,.2,4),locate(.3,.8,"V(0,0)"),
locate(-.2,4.35,"@"),locate(-.2,.35,"@"), line(-11,-4,11,-4) )}}}

Now draw a square with its left side as the line from the directrix
to the focus:

{{{drawing(400,400,-10,10,-10,10, locate(.3,4.3,"F(0,4)"),
rectangle(0,-4,8,4), locate(2,-4,DIRECTRIX),
graph(400,400,-10,10,-10,10), line(-.2,4,.2,4),locate(.3,.8,"V(0,0)"),
locate(-.2,4.35,"@"),locate(-.2,.35,"@"), line(-11,-4,11,-4),line(0,-4,0,4) )}}}

Now draw another square with its right side as the line from the directrix
to the focus:

{{{drawing(400,400,-10,10,-10,10, locate(.3,4.3,"F(0,4)"),
rectangle(0,-4,8,4), rectangle(0,-4,-8,4), locate(2,-4,DIRECTRIX),
graph(400,400,-10,10,-10,10), line(-.2,4,.2,4),locate(.3,.8,"V(0,0)"),
locate(-.2,4.35,"@"),locate(-.2,.35,"@"), line(-11,-4,11,-4),line(0,-4,0,4) )}}}

Finally sketch in the parabola passing through the 
upper far corners of the squares and through the vertex:

{{{drawing(400,400,-10,10,-10,10, locate(.3,4.3,"F(0,4)"),
rectangle(0,-4,8,4), rectangle(0,-4,-8,4), locate(2,-4,DIRECTRIX),
graph(400,400,-10,10,-10,10,x^2/16), line(-.2,4,.2,4),locate(.3,.8,"V(0,0)"),
locate(-.2,4.35,"@"),locate(-.2,.35,"@"), line(-11,-4,11,-4),
line(0,-4,0,4) )}}}

Edwin</pre>