Question 329448
Can you show me how to get the equation of a parabola with a focus (17,9) and a directrix x=12?
<pre><font size = 4 color = "indigo"><b>

The other tutor got the vertex wrong.

Let's draw the directrix line and the vertex:
{{{drawing(400,400,-2,20,-2,20,
locate(8.7,18,"x=12-->"),
graph(400,400,-2,20,-2,20), line(12,-10,12,30),
line(17+.2,9+.2,17-.2,9-.2), line(17+.2,9,17-.2,9),line(17,9+.2,17,9-.2),
locate(17,9,"F(17,9)")
 )}}}

The vertex is halfway between the focus and the directrix. 
We draw a green line from the focus to the directrix

{{{drawing(400,400,-2,20,-2,20,
locate(8.7,18,"x=12-->"),
graph(400,400,-2,20,-2,20), line(12,-10,12,30),
line(17+.2,9+.2,17-.2,9-.2), line(17+.2,9,17-.2,9),line(17,9+.2,17,9-.2),
locate(17,9,"F(17,9)"), green(line(17,9,12,9))
 )}}}

That green line is 5 units long, so the midpoint of the green line,
which is the vertex, is {{{2&1/2}}} units from the directrix and the 
focus.  So the coordinates of the vertex is ({{{14&1/2}}},9) or
({{{29/2}}},9)


{{{drawing(400,400,-2,20,-2,20,
locate(8.7,18,"x=12-->"),
graph(400,400,-2,20,-2,20), line(12,-10,12,30),
line(17+.2,9+.2,17-.2,9-.2), line(17+.2,9,17-.2,9),line(17,9+.2,17,9-.2),
locate(17,9,"F(17,9)"), green(line(17,9,12,9)),
locate(12.2,9,V(29/2,9)),
line(14.5+.2,9+.2,14.5-.2,9-.2), line(14.5+.2,9,14.5-.2,9),line(14.5,9+.2,14.4,9-.2)

 )}}}

To sketch in the parabola we construct two squares, one on each side of
the green line from vertex to focus:

{{{drawing(400,400,-2,20,-2,20,
locate(8.7,18,"x=12-->"),
graph(400,400,-2,20,-2,20), line(12,-10,12,30),
line(17+.2,9+.2,17-.2,9-.2), line(17+.2,9,17-.2,9),line(17,9+.2,17,9-.2),
locate(17,9,"F(17,9)"), green(line(17,9,12,9)),
locate(12.2,9,V(29/2,9)),

green(rectangle(12,4,17,14)),


line(14.5+.2,9+.2,14.5-.2,9-.2), line(14.5+.2,9,14.5-.2,9),line(14.5,9+.2,14.4,9-.2)

 )}}}

Now we can sketch in the parabola with the vertex and which passes
through the corners of those two squares:

{{{drawing(400,400,-2,20,-2,20,

graph(400,400,-2,20,-2,20), line(12,-10,12,30),
line(17+.2,9+.2,17-.2,9-.2), line(17+.2,9,17-.2,9),line(17,9+.2,17,9-.2),
locate(17,9,"F(17,9)"), green(line(17,9,12,9)),
locate(12.2,9,V(29/2,9)),
locate(8.7,18,"x=12-->"),
green(rectangle(12,4,17,14)),
graph(400,400,-2,20,-2,20,9+2sqrt(2.5(x-14.5))),graph(400,400,-2,20,-2,20,9-2sqrt(2.5(x-14.5))),

line(14.5+.2,9+.2,14.5-.2,9-.2), line(14.5+.2,9,14.5-.2,9),line(14.5,9+.2,14.4,9-.2)

 )}}}


The equation of the parabola which opens right or left and has vertex (h,k) 
is given by:

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

where p is the {{{2&1/2}}} or {{{5/2}}} unit distance between the 
directrix and the vertex, and also the same distance from the vertex 
to the focus.  p is taken positive if the parabola opens right, and 
negative if the parabola opens left.

This parabola opens right so {{{p=5/2}}}, and with the vertex
(h,k) = ({{{29/2}}},9)

{{{(y - 9)^2}}}{{{""=""}}}{{{4(5/2)(x - 29/2)}}}

{{{(y - 9)^2}}}{{{""=""}}}{{{10(x - 29/2)}}}

That's the equation of the parabola in standard form.

Edwin</pre>