Question 1041223
Question 1041196
<pre><b>
{{{(x+6)^2 = -24y+51}}}

We have to get it in the standard form:

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

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.  If p is positive
the parabola opens upward and if p is negative the
parabola opens downward.

The left side of

{{{(x+6)^2 = -24y+51}}}

is already in that form.  We factor out -24

{{{(x+6)^2=-24(y-51/24)}}}

Reduce the fraction:

{{{(x+6)^2=-24(y-17/8)}}}

Now we can compare it to

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

and see that 
-h=+6 so h=-6
-k = -17/8 so k = 17/8
4p = -24, so p = -6

The vertex is (h,k) or (-6,17/8).
We sketch the parabola. It goes through the 
vertex (-6,17/8) as opens downward since p 
is a negative number.

{{{drawing(400,400,-19,7,-13,13,
circle(-6,17/8,.15),
circle(-6,17/8,.2),
circle(-6,17/8,.25),
circle(-6,17/8,.3),

locate(-10.5,5,Vertex(matrix(1,3,-6,",",17/8))),
graph(400,400,-19,7,-13,13,((x+6)^2-51)/(-24))  )}}}
 
The focus (which is inside the parabola is |p| or 6
units below the vertex.  It has the same x-coordinate -6
So to find its y-coordinate we subtract 6 from the
y-coordinate of the vertex 17/8:

{{{17/8-6=17/8-48/8=-31/8}}}  

So the focus is (-6,-31/8)

{{{drawing(400,400,-19,7,-13,13,
circle(-6,17/8,.15),
circle(-6,17/8,.2),
circle(-6,17/8,.25),
circle(-6,17/8,.3),

locate(-10.5,5,Vertex(matrix(1,3,-6,",",17/8))),

circle(-6,-31/8,.15),
circle(-6,-31/8,.2),
circle(-6,-31/8,.25),
circle(-6,-31/8,.3),

locate(-10.5,-4,Focus(matrix(1,3,-6,",",-31/8))),

graph(400,400,-19,7,-13,13,((x+6)^2-51)/(-24)) )}}}

Finally we will draw the directrix, which is a horizontal
line (in green below) which is |p| = 6 units outside (above)
the vertex.  We determine how far above the x-axis that is
by adding 6 to the y-coordinate of the vertex 17/8:

{{{17/8+6=17/8+48/8=65/8}}}

So the directrix is a horizontal line y = 65/8

{{{drawing(400,400,-19,7,-13,13,
circle(-6,17/8,.15),
circle(-6,17/8,.2),
circle(-6,17/8,.25),
circle(-6,17/8,.3),
green(line(-25,65/8,10,65/8)),
locate(-10.5,5,Vertex(matrix(1,3,-6,",",17/8))),
locate(-10.5,9,Directrix(y=65/8)),
circle(-6,-31/8,.15),
circle(-6,-31/8,.2),
circle(-6,-31/8,.25),
circle(-6,-31/8,.3),

locate(-10.5,-4,Focus(matrix(1,3,-6,",",-31/3))),

graph(400,400,-19,7,-13,13,((x+6)^2-51)/(-24))  )}}}

  
The axis of symmetry is the vertical line (in blue below) 
through both the vertex and the focus which cuts the
parabola in two.  It is the line x=(the x-coordinate of
the vertex and focus).  In this case the equation of the
axis of symmetry is x=-6:

{{{drawing(400,400,-19,7,-13,13,
circle(-6,17/8,.15),
circle(-6,17/8,.2),
circle(-6,17/8,.25),
circle(-6,17/8,.3),
green(line(-25,65/8,10,65/8)),
locate(-10.5,5,Vertex(matrix(1,3,-6,",",17/8))),
locate(-10.5,9,Directrix(y=65/8)),
circle(-6,-31/8,.15),
circle(-6,-31/8,.2),
circle(-6,-31/8,.25),
circle(-6,-31/8,.3),
 blue(line(-6,17,-6,-18)),
locate(-10.5,-4,Focus(matrix(1,3,-6,",",-31/3))),

graph(400,400,-19,7,-13,13,((x+6)^2-51)/(-24))  )}}}
 
Edwin</pre>