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Question 688515: For the equation of the parabola find the coordinates for the VERTEX, FOCUS and the equations of the DIRECTRIX AND AXIS OF SYMMETRY
Y^2+6y+9=12-12x
Please help I've been working on it for an hour now.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! For the equation of the parabola find the coordinates for the VERTEX, FOCUS and the equations of the DIRECTRIX AND AXIS OF SYMMETRY
Y^2+6y+9=12-12x
(Y^2+6y+9)=12-12x-9+9
(y+3)^2=12-12x=12(1-x)=-12(x-1)
(y+3)^2=-12(x-1)
This is an equation of a parabola that opens leftwards
Its standard form: (y-k)^2=-4p(x-h), (h,k)=(x,y) coordinates of the vertex
For given parabola:
vertex: (1,-3)
axis of symmetry: y=-3
4p=12
p=3
focus: (-2,-3) (p-distance to the left of the vertex on the axis of symmetry)
directrix: x=4 (p-distance to the right of the vertex on the axis of symmetry)
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