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Question 621036: What are the coordinates of the vertex and foci, and what is the equation for the directrix of the following parabolic equations?
a. 
b. 
c. 
d. 
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! What are the coordinates of the vertex and foci, and what is the equation for the directrix of the following parabolic equations?
a.(x+2)^2 = -8*(y+3)
standard form of equation: (x-h)^2=-4p(y-k), (h,k)=(x,y) coordinates of vertex
parabola opens downwards (negative lead coefficient of right side)
vertex: (-2,-3)
axis of symmetry: x=-2
4p=8
p=2
focus:=(-2,-5) (p-units below vertex on axis of symmetry)
directrix: y=-1 (p-units above vertex on axis of symmetry)
..
b. (y-1)^2 = 16x
standard form of equation: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of vertex
parabola opens rightwards (positive lead coefficient of right side)
vertex: (0,1)
axis of symmetry: y=1
4p=16
p=4
focus:=(4,1) (p-units to the right of vertex on axis of symmetry)
directrix:x=-4 (p-units to the left of vertex on axis of symmetry)
..
c. x^2 = 4*(y-4)
standard form of equation: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of vertex
parabola opens upwards (positive lead coefficient of right side)
vertex: (0,4)
axis of symmetry: x=0
4p=4
p=1
focus:=(0,5) (p-units above vertex on axis of symmetry)
directrix: y=3 (p-units below vertex on axis of symmetry)
..
d. (y+6)^2 = -12*(x-1)
standard form of equation: (y-k)^2=-4p(x-h), (h,k)=(x,y) coordinates of vertex
parabola opens leftwards (negative lead coefficient of right side)
vertex: (1,-6)
axis of symmetry: y=-6
4p=12
p=3
focus:=(-2,-6) (p-units to the left of vertex on axis of symmetry)
directrix:x=4 (p-units to the right of vertex on axis of symmetry)
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