Question 1170110
<pre>
{{{drawing(500,500,-16,16,-6,26,
line(-6,3,6,3),



red(line(-sqrt(240),20,sqrt(240),20)),
graph(500,500,-16,16,-6,26,x^2/12))}}}

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

where (h,k) = (0,0) = the vertex and 4a = focal diameter

{{{matrix(1,4,
x^2,
""="",
(matrix(2,1,focal,diameter)),y))}}}

{{{matrix(1,3,
x^2,
""="",
12y)}}}

The black line is the focal diameter.

We want the length of the red line.

The end points are where y=20

{{{matrix(1,3,
x^2,
""="",
12y)}}}

{{{matrix(1,3,
x^2,
""="",
12(20))}}}

{{{matrix(1,3,
x^2,
""="",
240)}}}

{{{matrix(1,3,
x,
""="",
"" +- sqrt(240))}}}

{{{matrix(1,3,
x,
""="",
"" +- sqrt(16*15))}}}

{{{matrix(1,3,
x,
""="",
"" +- 4sqrt(15))}}}

So the red line is twice the absolute value of x.

Answer: {{{8sqrt(15)}}}, about 30.98

Edwin</pre>