Question 596993
Find the vertex,focus,and directrix of the equation; 
x^2+4x+6y-2=0
complete the square
(x^2+4x+4)+6y-2-4=0
(x+2)^2=-6y+6
(x+2)^2=-6(y-1)
This is an equation of a parabola that opens downwards of the standard form:
(x-h)^2=-4p(y-k), (h,k)=(x,y) coordinates of vertex
..
For given equation:
vertex:(-2,1)
axis of symmetry: x=-2
4p=6
p=3/2
focus:(-2,1-3/2)=(-2,1/2) (p units below vertex on axis of symmetry)
directrix: y=1+3/2
directrix: y=5/2 (p units above vertex on axis of symmetry)