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| Question 593014:  What is the focus, vertex, axis of symmetry and directrix of the parabola equation y2-8y+16x-64=0
 Answer by lwsshak3(11628)
      (Show Source): 
You can put this solution on YOUR website! What is the focus, vertex, axis of symmetry and directrix of the parabola equation y2-8y+16x-64=0
 complete the square
 (y^2-8y+16)+16x-64-16=0
 (y-4)^2+16x-80=0
 (y-4)^2=-16x+80
 (y-4)^2=-16(x-5)
 This is an equation for a parabola that open leftwards of the standard form: (y-k)^2=-4p(x-h), (h,k)=(x,y) coordinates of the vertex.
 For given parabola:
 4p=16
 p=4
 vertex: (5,4)
 axis of symmetry: y=4
 focus: (1,4) (p units to left of vertex on axis of symmetry)
 directrix: x=9 (p units to right of vertex on axis of symmetry)
 
 
 
 
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