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Question 783360: What is the vertex, focus and directrix of the following:
4y^2+4x+4y-3=0
2x^2-12X+3y+21=0
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! What is the vertex, focus and directrix of the following:
4y^2+4x+4y-3=0
4y^2+4y+4x-3=0
complete the square
4(y^2+y+1/4)-1+4x-3=0
4(y+1/2)^2=-4x+4
4(y+1/2)^2=-4(x-1)
divide by 4
(y+1/2)^2=-(x-1)
This is a parabola that opens leftward.
Its basic form of equation: (y+k)^2=-4p(x-h), (h,k)=(x,y) coordinates of the vertex
For given equation:
vertex: (1,-1/2)
axis of symmetry: y=-1/2
4p=1
p=1/4
focus: (3/4,-1/2) (p-distance left of vertex on the axis of symmetry
directrix: x=5/4 (p-distance right of vertex on the axis of symmetry
..
2x^2-12x+3y+21=0
complete the square
2(x^2-6x+9)-18+3y+21=0
2(x-3)^2=-3y+3
2(x-3)^2=-3(y+1)
(x-3)^2=-3/2(y+1)
This is a parabola that opens downward.
Its basic form of equation: (x-h)^2=-4p(y-k), (h,k)=(x,y) coordinates of the vertex
For given equation:
vertex: (3,-1)
axis of symmetry: x=3
4p=3/2
p=3/8
focus: (3,-11/8) (p-distance below vertex on the axis of symmetry
directrix: y=-5/8 (p-distance above vertex on the axis of symmetry
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