Question 621036
What are the coordinates of the vertex and foci, and what is the equation for the directrix of the following parabolic equations?
a.(x+2)^2 = -8*(y+3)
standard form of equation: (x-h)^2=-4p(y-k), (h,k)=(x,y) coordinates of vertex
parabola opens downwards (negative lead coefficient of right side)
vertex: (-2,-3)
axis of symmetry: x=-2
4p=8
p=2
focus:=(-2,-5) (p-units below vertex on axis of symmetry)
directrix: y=-1 (p-units above vertex on axis of symmetry)
..
b. (y-1)^2 = 16x
standard form of equation: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of vertex
parabola opens rightwards (positive lead coefficient of right side)
vertex: (0,1)
axis of symmetry: y=1
4p=16
p=4
focus:=(4,1) (p-units to the right of vertex on axis of symmetry)
directrix:x=-4 (p-units to the left of vertex on axis of symmetry)
.. 
c. x^2 = 4*(y-4)
standard form of equation: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of vertex
parabola opens upwards (positive lead coefficient of right side)
vertex: (0,4)
axis of symmetry: x=0
4p=4
p=1
focus:=(0,5) (p-units above vertex on axis of symmetry)
directrix: y=3 (p-units below vertex on axis of symmetry)
..

d. (y+6)^2 = -12*(x-1)
standard form of equation: (y-k)^2=-4p(x-h), (h,k)=(x,y) coordinates of vertex
parabola opens leftwards (negative lead coefficient of right side)
vertex: (1,-6)
axis of symmetry: y=-6
4p=12
p=3
focus:=(-2,-6) (p-units to the left of vertex on axis of symmetry)
directrix:x=4 (p-units to the right of vertex on axis of symmetry)