Answers

2014-06-12T11:44:41+05:30
This is an equation of a parabola of the standard form: (y-k)^2=4p(x-h), (h,k) being the (x,y) coordinates of the vertex. Parabola opens rightwards.
For given equation:
vertex: (-3,2)
4p=4
p=1
Focus=(-3+p,2)=(-3+1,2)=(-2,2) (p units from vertex on axis of symmetry)
Axis of symmetry: y=2 (a horizontal line thru the vertex)
Directrix: x=-4 (a vertical line p units from vertex on axis of symmetry)
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The Brainliest Answer!
2014-06-12T11:55:17+05:30
Y^2=4y - 4x 
=>y²=4(x-y)
=>y²+4y+4=4x-4
=>(y+2)²=4(x-1) which is the form of Y²=4aX
where X=x-1 and Y=y-2 and 4a=4,therefore a=1
For the vertex,
X=0,Y=0 i.e.,x+1=0,or x=1 and y-2=0,or y=2
Therefore coordinates of vertex are (1,2)
1 5 1