find the focus, the directrix and the equation of the axis of symmetry?

Question

1. y= 1/-32 y^2
2. x^2=-8y

Number 1. is actually x= 1/-32 y^2

Answer 1

1. x = 1/(4p) y², or y² = 4px
focus: (p, 0)
directrix: x = -p
axis of symmetry: y = 0

x = 1/(-32) y²
4p = -32
p = -8
focus: (-8, 0)
directrix: x = 8
axis of symmetry: y = 0

2. y = 1/(4p) x², or x² = 4py
focus: (0, p)
directrix: y = -p
axis of symmetry: x = 0

x² = -8y
4p = -8
p = -2
focus: (0, -2)
directrix: y = 2
axis of sym: x = 0

Answer 2

With y = x^2 / 20 fixing the equation for y = 0, the x fee is 0, giving the vertex (0,0). And x^2 = 20y reveals the concentration x^2 = 4py...p = 5 giving the concentration = (0.5) for this reason the "equidistant" directix is the line y = -5. With the x fee = 0 interior the vertex, the equation of the axis of symmetry will become x = 0.

Answer 3

Your 1st question makes no sense. You have copied the problem incorrectly

x^2= -8y
This is of the form x^2 = 2py , where p = -4
The vertex is at the origin.
The focus is at (0,p/2) = (0, -2)
The directrix is at y = -p/2 = 2
the axis of symmetry is the y-axis.

Answer 4

1. Did you mean x^2 instead of y^2?
2.Vertex form: y = 1/-8(x^2)
vertex: 0,0
a = 2 because 1/4a = -1/8
parabola opens down, so directrix is above and focus below
y=2 directrix
focus below (0,-2)
axis of symmetry x=0 because x-coordinate of vertex 0

Answer 5

x² = -8y
y = (-1/8)x²

Take the first derivative:
y' = -(1/4)x
y' = 0   ⇒   x = 0
axis of symmetry: x = 0

Answer 6

Make sure that
f(2a-x)=2b-f(x)
where a and b are the coordinates of the focus .