Question 415344
Find the equation for a parabola with its focus at (5,0) and with 
directrix x=-5. 
<pre><font face = "DotumChe" color = "indigo" size = 4><b>
Here is its focus, the point F(5,0), and its directrix, the green
line below.

{{{drawing(600,600,-15,15,-15,15, graph(600,600,-15,15,-15,15),
locate(5.1,1,"F(5,0)"),
circle(5,0,.16), green(line(-5,-16,-5,16))  )}}}


The vertex of a parabola is half-way between the focus point and the 
directrix line.  Looking at the graph, that puts the vertex at the origin
V(0,0).  p is the distance from the vertex to the focus, taken as a positive
number if the parabola is to open to the right and p is taken as a negative 
number if the parabola is to open to the left.

{{{drawing(600,600,-15,15,-15,15, graph(600,600,-15,15,-15,15),
locate(5.1,1,"F(5,0)"), locate(-2.8,1,"V(0,0)"),
circle(5,0,.16), green(line(-5,-16,-5,16))  )}}}







This one opens to the right, so p is taken positive as 5, the distance
between the vertex and the focus, which is also half of the distance between
the focus and the directrix.  So its equation is

(y - k)² = 4p(x - h)

and since the vertex is (h,k) = (0,0) and p = 5, that simplifies to

(y - 0)² = 4*5(x - 0)

      y² = 20x 

That's the equation you wanted

To finish the graph we constract a square on each side of the line
that goes from the focus through the vertex to the directrix, like this:

{{{drawing(600,600,-15,15,-15,15, graph(600,600,-15,15,-15,15),
locate(5.1,1,"F(5,0)"), locate(-2.8,1,"V(0,0)"),
circle(5,0,.16), green(line(-5,-16,-5,16)), rectangle(-5,-10,5,10)  )}}}

Then we can sketch in the parabola through the vertex and the right 
uppermost and lowermost corners of those two squares:

{{{drawing(600,600,-15,15,-15,15, graph(600,600,-15,15,-15,15,-sqrt(20x)),
locate(5.1,1,"F(5,0)"), locate(-2.8,1,"V(0,0)"),
graph(600,600,-15,15,-15,15,sqrt(20x)),circle(5,0,.16), green(line(-5,-16,-5,16)), rectangle(-5,-10,5,10)  )}}}

Edwin</pre>