SOLUTION: Find the equation in standard form of the given parabola as a conic section. Focus is at (-2, 5) Directrix is y=1

That will either be 

(x-h)² = 4p(y-k), if the parabola is like a U or an upside down U.

or it will be

(y-k)² = 4p(x-h), if the parabola is like a C or a backwards C.

where (h,k) is the vertex and p is the distance from the vertex
to the focus and also the distance from the vertex to the
directrix.  p is taken positive is like a U or a C, and negative
if it is like an upside-down U or a backwards C.

First we will sketch the graph:

Plot the focus and graph the directrix:

  

Draw a line from the focus straight down through the vertex
all the way to the directrix: 

  

Now sketchin the parabola through the vertex and the upper
outer corners of those two squares, like this:


So the parabola is like a U so it has the form:

(x-h)² = 4p(y-k)  and p is taken positive.

The vertex is V(h,k) =  V(-2,3) and p is the distance from the vertex to
the focus and also the distance from the vertex to the
directrix which is p=2, so substitute h=-2,k=3,p=2:

(x-(-2))² = 4(2)(y-3)

(x+2)² = 8(y-3)

Edwin