Question

Help please,

Find the vertex,focus and directrix of the following parabola. (x+3)^2=-3(y-2)

Answer 1

(x+3)^2=-3(y-2)
(x-(-3))^2 = -3y + 6
3y = -1*(x-(-3))^2 + 6

Answer 2

I'm just trying to get it into the regular form:
y = (-1/3)*(x-(-3))^2 + 2

Answer 3

okay...i got (-3,2) for vertex
for focus i got (-3,-3)
and for directrix i got -1

are these correct?

Answer 4

y-2 = (-1/3)*(x-(-3))^2
vertex is (h,k), so in this case: (-3, 2). yes vertex is right

Answer 5

a = -1/3, h = -3, k = 2

Answer 6

focus is (a + h, k)
((-1/3)-3, 2) = (-10/3, 2)

Answer 7

okay i see

Answer 8

what was the formula for directrix? amistre am i doing this right?

Answer 9

I dont remember any of this, im looking up formulas that might be wrong.

Answer 10

directix is the opposite direction, same distance, as focus

Answer 11

y^2 = 4ax is a standard parabola ; where "a" measure is the distance from vertex to focus/directix

Answer 12

(x+3)^2=-3(y-2)

x^2 = -3y is our parabola centerd at the origin so that we can see it better

-3 = 4a

a = -3/4

Answer 13

so that is the directrix ?..also does the mean the focus was wrong?

Answer 14

*this

Answer 15

it means that the distance to our focus is -3/4 from the vertex; and the directix is in the other direction

Answer 16

ahhh okay i understand the picture helped

Answer 17

our vertex is at -3,2 right?

Answer 18

-3,-6 might be better

Answer 19

(x+3)^2 = -3y + 6

(x+3)^2 - 6= -3y

-1/3 (x+3)^2 - 6= y

yeah, vertex = -3,-6

our focus is then at (-3,(-6- 3/4)); and the directix is the horisontal line at y=(-6+3/4)

Answer 20

almost did that right lol

Answer 21

yes..okay i understand you..makes sense now thanks..amistre and bahrom! :D

Answer 22

if you can find my mistake, and correct it, itll prolly be helpful

Answer 23

at any rate, click on properties and itll help you chk your answers

http://www.wolframalpha.com/input/?i=%28x%2B3%29%5E2%3D-3%28y-2%29

Answer 24

Okay will do thanks