Question 599466: What is the vertex, focus, axis of symmetry and focal diameter of the equation 9y^2+36x+18y-63=0
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! What is the vertex, focus, axis of symmetry and focal diameter of the equation
9y^2+36x+18y-63=0
complete the square
9(y^2+2y+1)+36x-63-9=0
9(y+1)^2=-36x+72=-36(x-2)
(y+1)^2=-4(x-2)
This is an equation of a parabola which opens leftwards.
Its standard form: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of vertex
For given equation:
vertex: (2,-1)
axis of symmetry: y=-1
4p=4
p=1
focus: (1,-1) (p distance from vertex on the axis of symmetry)
focal diameter=4p=4
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