SOLUTION: 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?

Let's draw the axis of symmetry, 

 

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

Now we'll draw the directrix in green:

 

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



(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 .

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




The vertex is p or  units above the directrix.

Therefore the vertex is (-2,5) or (-2,)

and the equation is





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

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

 

Now the focus is  units above the
vertex, so we plot it at (-2,7)



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:


 
Now we can sketch in the parabola through the vertex and the 
end-points of the latus rectum:
 


Edwin