SOLUTION: Find the vertex,focus,directrix,and the axis of symmetry in the given equation: (x+6)^2 = -24y+51

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Question 1041223: Find the vertex,focus,directrix,and the axis of symmetry in the given equation:
(x+6)^2 = -24y+51

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Question 1041196

%28x%2B6%29%5E2+=+-24y%2B51

We have to get it in the standard form:

x-h%29%5E2=4p%28y-k%29

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

%28x%2B6%29%5E2+=+-24y%2B51

is already in that form.  We factor out -24

%28x%2B6%29%5E2=-24%28y-51%2F24%29

Reduce the fraction:

%28x%2B6%29%5E2=-24%28y-17%2F8%29

Now we can compare it to

x-h%29%5E2=4p%28y-k%29

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.


 
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%2F8-6=17%2F8-48%2F8=-31%2F8  

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



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%2F8%2B6=17%2F8%2B48%2F8=65%2F8

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



  
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:


 
Edwin