SOLUTION: Find the vertex, focus and directrix and sketch the graph of the parabola from the equation (x+1)^2 + 8(y+2)=0. First, I got the vertex (-1,-2), h=-1 k=-2. Next I tried to find

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Question 691514: Find the vertex, focus and directrix and sketch the graph of the parabola from the equation (x+1)^2 + 8(y+2)=0.
First, I got the vertex (-1,-2), h=-1 k=-2. Next I tried to find teh focus but wanted to get "p" first. So 4p=8 p=2
Using the formula to get the focus (h,k+p) I get (-1,-2+2)-(1,0) but when I looked at my answer book it said that p=-2 and the focus is (-1,-4). What may I be doing wrong.

Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
The correct standard form of theequation of a parabola with
a vertical axis of symmetry is

(x - h)² = 4p(y - k)

Your parabola equation, 

(x + 1)² + 8(y + 2) = 0

is not quite in that form.  So we subtract 8(y + 2)
from both sides:

           (x + 1)² = -8(y + 2)

Compare that to

           (x - h)² = 4p(y - k) 

And we see that h = -1, k = -2 and 4p = -8
                                    p = -2

We plot the vertex (h,k) = (-1,-2), and since p = -2, a
negative number, the focus is 2 units BELOW the vertex. 
So the focus is (-1,-4).  And the directrix is 2
units ABOVE the vertex and therefore has equation y = 0,
which just happens to be the x-axis.  The focal chord or
latin rectum is a line through the focus, bisected by the
focus ad which is |4p| = 8 units long.  So we draw in the latus
rectum, in green: 
 



and sketch in the parabola:



Edwin

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