Finding the vertex, focus, and directrix of a parabola?

Question

I need to find the Focus and directrix of a parabola.... I have the answer, but I need to know how to get to the answers.

y^2+6y+8x+25 = 0

Answer 1

Put the equation of the parabola into the form

4p(x - h) = (y - k)²

The vertex is (h,k) and the distance from the vertex to the focus is p. The distance from the vertex to the directrix is also p.

For the question at hand we have:

y² + 6y + 8x + 25 = 0
8x + 25 = -y² - 6y
8x + 25 - 9 = -(y² + 6y + 9)
8x + 16 = -(y + 3)²
-8(x + 2) = (y + 3)²

The vertex (h,k) is (-2,-3).

Since the y term is squared the parabola opens sideways. Since the y squared term and the x term have opposite signs and are on the opposite side of the equal sign, the parabola opens to the left.

The line of symmetry is a horizontal line thru the vertex.
y = -3

4p = -8
p = -2

The focus is a distance | p | = 2 from the vertex along the line of symmetry. It is inside the parabola. The focus is

(h + p, k) = (-2 - 2, -3) = (-4, -3)

The directrix is also a distance | p | = 2 from the vertex but in the opposite direction. It is a line perpendicular to the line of symmetry and is outside the parabola. The directrix is a vertical line with the equation:

x = h - p = -2 + 2 = 0
x = 0

Answer 2

y^2+6y+8x+25 = 0
Complete the square in y:
y^2 +6y +9 +8x +25 = 9
(y+3)^2 = -8x-16= -8(x+1)
This is the equation of a parabola with axis of symmetry parallel to the x-axis.
Its vertex is at (-1,-3)
2p= - 8 so p = -4 So the focus -p/2 = 2 units to the left of the vertex and the directrix is 2 units to the right of the vertex.

Thus the focus is at (-3, -3) and the equation of the directrix is
x = 1.