SOLUTION: Convert the equation to standard form by completing the square on x or y. Then find the vertex, focus? and directrix of the parabola. Please show work so I know you got your a

Question 154307: Convert the equation to standard form by completing the square on x or y. Then find the vertex, focus?
and directrix of the parabola.
Please show work so I know you got your answer and I can learn, please.

1. x^2+6x+8y+1=0
2. y^2-2y-8x+1=0
Found 2 solutions by scott8148, terasse jade:
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
one general form of the equation of a parabola is (x-h)^2=4p(y-k)
__ (the x and y may be switched, depending on the direction of the parabola)
__ the vertex is (h,k), and p is the distance from the vertex to the focus
__ since the vertex is midway between the focus and directrix, -p is the distance from the vertex to the directrix
__ the vertex and focus are on the axis of symmerty, the directrix is perpendicular to the axis

so, the "trick" is to manipulate the equation until it looks like the general form

x^2+6x+8y+1=0 __ subtracting 8y+1 __ x^2+6x=-8y-1 __ completing the square by adding (6/2)^2 (from 6x) __ x^2+6x+9=-8y-1+9

x^2+6x+9=-8y+8 __ factoring __ (x+3)^2=-8(y-1)
__ vertex is (-3,1)
__ focus is (-3,-1)
__ directrix is y=3

y^2-2y-8x+1=0 __ subtracting -8x+1 __ y^2-2y=8x-1 __ completing the square by adding (-2/2)^2 __ y^2-2y+1=8x-1+1

y^2-2x+1=8x __ factoring __ (y-1)^2=8(x-0)
__ vertex is (0,1)
__ focus is (2,1)
__ directrix is x=-2

Answer by terasse jade(1) (Show Source): You can put this solution on YOUR website!
13

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