Finding vertex, focus and directrix of parabola this given equation:?

Question

y^2 + 2y - 8x -7 =0

Answer 1

y^2 + 2y + 1 = 8x + 7 + 1

(y+1)^2 = 8x + 8

8(x+1) = (y+1)^2

8 = 4c , distance from vertex to focus is c=2. Parabola opens right since y is squared and 8 is positive.

Vertex (-1, -1)

Focus: (1, -1) (2 places right of vertex)

directrix: x=-3 (2 places left of vertex, vertical line)

Answer 2

Given the parabola in general form:
y² + 2y - 8x - 7 = 0

You must convert it into the form
(y - k)² = 4a(x - h)
Where (h,k) is the vertex, (h + a,k) is the focus and the line x = h - a is the directrix.

Again:
y² + 2y - 8x - 7 = 0

Transpose the x term and the constant
y² + 2y = 8x + 7

Add 1 to both sides (so we can perfect the trinomial square at the left)
y² + 2y + 1 = 8x + 7 + 1

Factor:
(y + 1)² = 8(x + 1)
(y - k)² = 4a(x - h)
Therefore,
-k = 1
k = -1,

-h = 1
h = -1,
4a = 8
a = 2.

The vertex is
(h,k) = (-1, -1)
The focus is
(h + a,k) = (1,-1)
The directrix is
x = h - a
x = -3

^_^

Answer 3

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