Focus, Directrix, and axis of symmetry for parabolas?

Question

heres the original equation: x=1/4y^2+2y-2. so i think the equation in graphing form is x=(y+4)^2-2. this would make the vertex (-2,-4). i think. but i dont know how to find the focus, directrix, or axis of symmetry. can somebody please help? i need to be able to graph this.

but wait since the parabola is supposed to look like a "< ", wouldn't the Axis of symmetry be Y=-4 and the DIRECTRIX be X=-4 ?????????

Answer 1

Ans:
This is a parabola open apward and shifted to -4 in the y-axis
Its vertex is (-2,-4)
Focus is on the y-axis
Therefore it is (0,-4)
Directrix is y=-4
Axis of symmetry is the Y-axis

Answer 2

x = 1/4y^2 + 2y - 2
x = 1/4(y^2 + 8y - 8)
x = 1/4(y^2 + 8y + 16 - 16 - 8)
x = 1/4((y^2 + 8y + 16) - 24)
x = 1/4(y+ 4)^2 - 6
x + 6 = 1/4(y+ 4)^2

The vertex is at (-6, -4),
making the axis of symmetry y = -4
(x - k) = (1/(4p)) (y - h)^2
1/(4p) ≡ 1/4, p = 1
Focus is at (- 5,-4)
Directrix is x = - 7

Answer 3

x = a million/32 y^2 y^2 = 32x So 2p = 32 --> p = sixteen The vertex is on the commencing place Its axis is the x-axis the concentration is at (0, p/2) = (0,8) The equation of the directrix is x = 8 2p = 32 is the size of the latus rectum that's the size of the line during the concentration and || to axis of symmetry.