SOLUTION: im familar with writing conic sections in there standard equation but there are 2 questions that have me stumped in peticular.here is one of them. y^2-8y-8x+64=0 and im suppose

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: im familar with writing conic sections in there standard equation but there are 2 questions that have me stumped in peticular.here is one of them. y^2-8y-8x+64=0 and im suppose      Log On


   

Question 170136: im familar with writing conic sections in there standard equation but there are 2 questions that have me stumped in peticular.here is one of them.
y^2-8y-8x+64=0
and im suppose to write this in its standard equation.
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
this is a parabola where the x and y have been "swaped"

the axis of symmetry is horizontal instead of vertical __ parabola opens to the right (positive x)

adding 8x-64 __ y^2-8y=8x-64

completing the square __ y^2-8y+16=8x-48

multiplying by 1/8 __ (1/8)(y-4)^2=x-6

adding 6 __ (1/8)(y-4)^2 + 6 = x __ this is the vertex form

standard form __ x = (1/8)y^2 - y + 8