Question

ind the vertex, value of p, axis of symmetry, focus, and directrix of the parabola.

y−2=1/12(x−4)^2

Answer 1

@gorv

Answer 2

y-y1=(1/12)(x-x1)^2

Answer 3

vertex
(x1,y1)

Answer 4

what is vertex ??

Answer 5

line of symmetry
x=x1

Answer 6

at @fani1996

Answer 7

hmmm (4,2)

Answer 8

focus( 4,5)

Answer 9

@ganeshie8

Answer 10

do you have general equation of a parabola with vertex

Answer 11

idk im confused

Answer 12

General form of parabola with vertex (h,k), there are two cases:

(x-h)^2 = 4p (y - k )

or

(y-k)^2 = 4p(x-h)

Answer 13

in your problem, the square is on the (x-h) , so we will use the first form

Answer 14

hmm ok

Answer 15

the axis of symmerty x= -1?

Answer 16

General form of parabola with vertex (h,k).There are two cases:

Case 1: Parabola opens upwards or downwards
equation: (x-h)^2 = 4p (y - k )
focus: (h,k+p)
directrix : y = k-p
Axis : x = h
--------------------------------------------------------
Case 2: Parabola opens to the right or left (sideways)
equation: (y-k)^2 = 4p(x-h)
Focus: (h+p, k)
Directrix: x = h- p
Axis : y = k

Answer 17

are my answers right

Answer 18

do i have the focus and vertex

Answer 19

Given problem : 1/12(x-4)^2 = y-2

Get into form above

(x-4)^2 = 12 (y-2)

now it is in form (x-h)^2 = 4p(y-k)

h = 4, k = 2 ,
4p = 12 ==> p = 3

Answer 20

ok so far?

Answer 21

hmm yes i had p=3

Answer 22

focus: (4,2+3)
directrix : y = 2-3
Axis : x = 3

Answer 23

diretrix : y = -1

Answer 24

axis of symmetry x= 4

Answer 25

ahh so i had the vertex and focus and p= 3 right thanks for helping me and i understand better