SOLUTION: I need to find the vertex, focus, directrix and axis of symmetry of each parabola 6x+y^2=0 I know (y+0)^2 is part of a formula but what do I do with the 6x since there is no sq

Question 674016: I need to find the vertex, focus, directrix and axis of symmetry of each parabola
6x+y^2=0
I know (y+0)^2 is part of a formula but what do I do with the 6x since there is no square?
another problem is x^2=y+2x Can you tell me how to set this one up?
Thank you!
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
I need to find the vertex, focus, directrix and axis of symmetry of each parabola
6x+y^2=0
y^2=-6x
This is an equation of a parabola that open leftwards.
Its standard form: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of the vertex
For given equation: y^2=-6x
vertex:(0,0)
axis of symmetry: y=0
4p=6
p=6/4=3/2
focus: (-3/2,0) (p-distance left of vertex on axis of symmetry
directrix: x=3 (p-distance right of vertex on axis of symmetry)
..
x^2=y+2x
y=x^2-2x
This is an equation of a parabola that open upwards.
Its standard form:y=A(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex, A>0, curve opens upwards, A<0, curve opens downwards.
for given equation: y=x^2-2x
complete the square
y=(x^2-2x+1)-1
y=(x-1)^2-1
rewrite to standard form of first equation above
(x-1)^2=(y+1)
vertex: (1,-1)
axis of symmetry: x=1
4p=1
p=1/4
focus: (1,-3/4) (p-distance above vertex on axis of symmetry)
directrix: y=-5/4 (p-distance below vertex on axis of symmetry)

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