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Question 621035: 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)
This is an equation of a parabola that opens downwards
Form of equation: (x-h)^2=-4p(y-k)
vertex:(-2,-3)
axis of symmetry: x=-2
4p=8
p=2
focus: (-2,-3-p)=(-2,-3-2)=(-2,-5) (p units below vertex on the axis of symmetry)
directrix: y=-1 (p units above vertex on the axis of symmetry)
..
b. (y-1)^2 = 16x
This is an equation of a parabola that opens rightwards
Form of equation: (y-k)^2=4p(x-h)
vertex:(0,1)
axis of symmetry: y=1
4p=16
p=4
focus: (0+p,1)=(0+4,1)=(4,1) (p units right of vertex on the axis of symmetry)
directrix: x=-4 (p units left of vertex on the axis of symmetry)
..
c. x^2 = 4*(y-4)
This is an equation of a parabola that opens upwards
Form of equation: (x-h)^2=4p(y-k)
vertex:(0,4)
axis of symmetry: x=0
4p=4
p=2
focus: (0,4+p)=(0,4+2)=(0,6) (p units above vertex on the axis of symmetry)
directrix: y=2 (p units below vertex on the axis of symmetry)
..
d. (y+6)^2 = -12*(x-1)
This is an equation of a parabola that opens leftwards
Form of equation: (y-k)^2=-4p(x-h)
vertex:(1,-6)
axis of symmetry: y=-6
4p=12
p=3
focus: (1-p,-6)=(1- 3,-6)=(-2,-6) (p units left of vertex on the axis of symmetry)
directrix: x=4 (p units right of vertex on the axis of symmetry)
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