Question

Identify the vertex, focus, and directrix of the parabola with the equation x^2-6x-8y+49=0

Answer 1

Get it in form 4P(y-k)=(x-h)^2
add 8y to both sides
x^2-6x+49=8y
complete the square
-6/3=-3, (-3)^2=9
(x^2-6x+9)+40=8y
(x-3)^2+40=8y
minus 40 both sides
(x-3)^2=8y-40
undistribute 8 on right side
(x-3)^2=(y-5)8
4P(y-k)=(x-h)^2
4(2)(y-5)=(x-3)^2

(h,k) is center
p is distance from vertex to directix and also from vertex to focus
p=2
vertex=(3,5)
in that form, the prabola opens up so
directix is 2 below vertex
focus is 2 above vertex
up and down is y so
5-2=3
5+2=7

directix is y=3
focus=(3,7)
vertex=(3,5)

Answer 2

Simple...

Graph the parabola using the direction, vertex, focus, and axis of symmetry...

Direction:Opens Up

Vertex:(3,5)

Focus:(3,7)

Axis of Symmetry:x=3

Directrix:y=3

A table of coordinates...

   x       |            y
   1       |         5.5
   2       |         5.125
   3       |         5         
   4       |         5.125
   5       |         5
            |
            |
            |