Question 154471
Find equation of parabola whose axis is x+2=0,
directrix is y-4=0, latus rectum=6 and concativity 
towards positive direction of y-axis?
<pre><font size = 4 color = "indigo"><b>
Let's draw the axis of symmetry, {{{x+2=0}}}

{{{drawing(400,400,-7,5,-2,10, 
graph(400,400,-7,5,-2,10), rectangle(-8,-3,-2,12) )}}} 

The vertex and the focus must be on this line of
symmetry.  So their x-coordinates must be {{{-2}}}

Now we'll draw the directrix in green:

{{{drawing(400,400,-7,5,-2,10, 
graph(400,400,-7,5,-2,10,0,4), rectangle(-8,-3,-2,12) )}}} 

The equation of a parabola which is concave upward
or downward is

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

(h,k) is the vertex

p = directed distance from directrix to vertex = 
distance from vertex to focus.

Here we use the fact that the latus rectum
has length {{{abs(4p)}}}.

We are told the the latus rectum has length 6.
and p is positive since parabola is concave 
upward, so

{{{4p = 6}}}
{{{p=6/4 = 3/2}}}

The vertex is p or {{{3/2}}} units above the directrix.

Therefore the vertex is (-2,5{{{1/2}}}) or (-2,{{{11/2}}})

and the equation is

{{{(x-(-2))^2=4(3/2)(y-(11/2))}}}

{{{(x+2)^2=6(y-11/2)}}}

That is all you wanted. But let's plot the
other parts and draw the graph.

Let's plot the vertex with an "o",

{{{drawing(400,400,-7,5,-2,10, locate(-2-.1,11/2+.3,o), 
graph(400,400,-7,5,-2,10,0,4), rectangle(-8,-3,-2,12) )}}} 

Now the focus is {{{P=1}}}{{{1/2}}} units above the
vertex, so we plot it at (-2,7)

{{{drawing(400,400,-7,5,-2,10, locate(-2-.1,11/2+.3,o), locate(-2-.1,7+.3,o),
graph(400,400,-7,5,-2,10,0,4), rectangle(-8,-3,-2,12) )}}}

Let's draw the latus rectum:

The latus rectum is the horizontal line segment the ends
of which are on the parabola.  Since the latus rectum is 
6 units long, we draw it 3 units right of the focus and
3 units left of the focus:

{{{drawing(400,400,-7,5,-2,10, locate(-2-.1,11/2+.3,o), locate(-2-.1,7+.3,o),
graph(400,400,-7,5,-2,10,0,4), line(-5,7,1,7), rectangle(-8,-3,-2,12) )}}}
 
Now we can sketch in the parabola through the vertex and the 
end-points of the latus rectum:
 
{{{drawing(400,400,-7,5,-2,10, locate(-2-.1,11/2+.3,o), locate(-2-.1,7+.3,o),
graph(400,400,-7,5,-2,10,(x+2)^2/6+11/2,4), line(-5,7,1,7), rectangle(-8,-3,-2,12) )}}}

Edwin</pre>