Question 768972
Focus at (2, 4); directrix the line x = -4. Find the equation of the parabola described and two point that define the latus rectum. 
<pre>

The other tutor's answer is incorrect.

Since the directrix is vertical, the parabola has the equation

(y - k)² = 4p(x - h)

where (h,k) is the vertex, and p is the distance from the vertex
to the focus, taken positive if the parabola opens right and
negative if the parabola opens left.  We can tell that the 
parabola opens to the right, since the directrix is outside the
parabola and the focus is inside the parabola.  

We plot the focus and draw the directrix line in green:

{{{drawing(300,400,-7,8,-6,14,graph(300,400,-7,8,-6,14),grid(1),

circle(2,4,.2), green(line(-4,-15,-4,20)) )}}}

The vertex is the midpoint of the line segment drawn from the
focus perpendicular to the directrix, drawn in red, and its midpoint
which is the vertex is the point (h,k) = (-1,4).  So the equation
so far 

(y - k)² = 4p(x - h)

becomes

(y - 4)² = 4p(x - (-1))

(y - 4)² = 4p(x + 1)

{{{drawing(300,400,-7,8,-6,14,graph(300,400,-7,8,-6,14),grid(1),
red(line(2,4,-4,4)),circle(-1,4,.2),
circle(2,4,.2), green(line(-4,-15,-4,20)) )}}}

So the distance p = +3, the distance from the vertex to the focus.
p is taken positive because the parabola opens to the right.  So 
the equation is

(y - 4)² = 4(3)(x + 1)

(y - 4)² = 12(x + 1)

The latus rectum is the line 4p = 12 units long, parallel to the
directrix and its midpoint is the focus.  So we draw the latus
rectum in blue:

{{{drawing(300,400,-7,8,-6,14,graph(300,400,-7,8,-6,14),grid(1),
red(line(2,4,-4,4)),circle(-1,4,.2), blue(line(2,-2,2,10)),
circle(2,4,.2), green(line(-4,-15,-4,20)) )}}}

And the two points that define the latus ractum are its end points
(2,-2), and (2,10), and here is the graph:

{{{drawing(300,400,-7,8,-6,14,graph(300,400,-7,8,-6,14,4+sqrt(12(x+1))),grid(1),
red(line(2,4,-4,4)),circle(-1,4,.2), blue(line(2,-2,2,10)),
graph(300,400,-7,8,-6,14,4-sqrt(12(x+1))),red(line(-1,3,-1,5)),
circle(2,4,.2), green(line(-4,-15,-4,20)) )}}}

Edwin</pre>