I nee help the help in finding vertex, focus, and directrix of the follwoing parabola,

3x+y^2+8y+4=0

I get -52/3 vertex, -4,-52/3 and directrix -100/3/ I am not gettind right according to my teacher and the book. Am I even close, Thanks for any answers, help

2. Originally Posted by creatureman
I nee help the help in finding vertex, focus, and directrix of the follwoing parabola,

3x+y^2+8y+4=0

I get -52/3 vertex, -4,-52/3 and directrix -100/3/ I am not gettind right according to my teacher and the book. Am I even close, Thanks for any answers, help
3x + y^2 + 8y + 4 = 0

3x + 4 = -(y^2 + 8y)

Now complete the square:
3x + 4 = -(y^2 + 8y + 16 - 16)

3x + 4 = -(y^2 + 8y + 16) + 16

3x - 12 = -(y + 4)^2

Now factor out a 3.

3(x - 4) = -(y + 4)^2

Now write this as
4*(3/4)(x - 4) = -(y + 4)^2

Compare this to the form of a leftward opening parabola with a vertex at (h, k):
4p(x - h) = -(y - k)^2
where p is the distance from the vertex to the focus, and the distance from the focus to the directrix.

So p = 3/4, and the vertex is at (4, -4).

The focus is a point p = 3/4 units to the left of the vertex:
(4 - 3/4, -4) = (13/4, -4)

And the directrix is a vertical line p = 3/4 units to the right of this point:
x = 4 + 3/4 = 19/4

-Dan