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

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

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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)