Question

whats the focus of the parabola y=1/8(x-2)^2 -4

Answer 1

can anyone help me!?

Answer 2

it we put it into the geometric form of: (y-k)^2 = 4a(x-h) itll be easier to deal with

Answer 3

how do you do that?

Answer 4

by moving things about :)

Answer 5

but what are all those variables?

Answer 6

they define the center, and the distance from the vertex to the focus/directix

Answer 7

y=1/8(x-2)^2 -4
+4 +4

y+4 =1/8(x-2)^2
*8 *8

8(y+4) = (x-2)^2

(4*2) (y+4) = (x-2)^2

Answer 8

trying to remember if i sholda moved the 8 :/

Answer 9

ohh okay! so then whats the focus?

Answer 10

(y+4) =(1/8) (x-2)^2

(y+4) = 4 (1/32) (x-2)^2

that might be better

Answer 11

the center and vertex of a parabola are the same point; (2,-4) if we read it form the equation

Answer 12

the focus is "a" away from the center in the direction of the open part; in this case add 1/32 to the center y

Answer 13

and to make sure im remembering it right, well chk with the wolf

Answer 14

http://www.wolframalpha.com/input/?i=y%3D1%2F8%28x-2%29%5E2+-4

pfft, i was right the first time with a=2 .....

Answer 15

So that would make the focus (2, -2) right?

Answer 16

correct

Answer 17

yay! thankyou!

Answer 18

to recap the correct rendition :)

y=1/8(x-2)^2 -4
+4 +4

y+4 =1/8(x-2)^2
*8 *8

8(y+4) = (x-2)^2

4(2) (y+4) = (x-2)^2
4 a (y -k) = (x-h)^2

center is (h,k), and we add "a"

good luck