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Question 596993: Find the vertex,focus,and directrix of the equation; x^2+4x+6y-2=0
(What I've tried)
X^2+4x+6y-2=0
(x+2)^2 = 6y+2 (here's where i'm confusing myself)
(x+2)^2 = 6(y+2)? If that's right then it should be...
(x+2)^2 = 4(6/4)(y+2)
P=(6/4)
Vertex = (-2,-2)
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! 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)
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