determine  the focus, vertex and directrix of the parabola of x^2+6x-8y+17=0
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x²+6x-8y+17=0,

x²+6x+9-9-8y+17=0,

(x+3)²-9-8y+17=0,

(x+3)²-8y+8=0,

(x+3)²/8-y+1=0,

y-1=(x+3)²/8, so vertex is at (-3,1).

The coefficient of the squared term is 1/8=1/4f where f is the vertical distance of the focus and directrix line from the vertex, so 4f=8, f=2. The focus is “inside” the parabola, so y=1 at the vertex, and the focus is at y=1+2=3 on the axis of symmetry x=-3, so its coords are (-3,3).

The directrix is “outside” the parabola at y=1-2=-1, the line y=-1.

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