Question

how do i find the properties of this hyperbola -4x^2+y^2-8x-12y+36=0

Answer 1

generally by completing the squares for the x and y parts

Answer 2

can you show me how to work the problem out

Answer 3

yes, but it requires knowledge of completing a square; do you remember how to work that?

Answer 4

yes little bit i would have to group my common factors first right?

Answer 5

common x and y parts yes

-4x^2+y^2-8x-12y+36=0

(-4x^2-8x + ___ ) + (y^2-12y + ___ ) +36 = 0

Answer 6

looks like the 36 already completes the y square for you

Answer 7

(-4x^2-8x + ___ ) + (y^2-12y + 36) = 0

(-4x^2-8x + ___ ) + (y-6)^2 = 0

Answer 8

-24?

Answer 9

dunno, lets factor out that -4 from the x parts

(-4x^2-8x + ___ ) + (y-6)^2 = 0

-4 (x^2+2x + ___ ) + (y-6)^2 = 0

id say +1 -1

-4 (x^2+2x + 1 - 1) + (y-6)^2 = 0

-4 (x^2+2x + 1) +4 + (y-6)^2 = 0

-4 (x+1)^2 + (y-6)^2 = -4

Answer 10

divide both sides by -4 and we have ...

(x+1)^2 (y-6)^2
------- - ------- = 1
1 4

Answer 11

now its starting to come back to me

Answer 12

after this step i get lost

Answer 13

well, we can determine alot of properties from this now; it all depends what properties you need to find

Answer 14

foci

Answer 15

well, we would need to determine which way this opens up; since -y^2 can never be positive, it opens along the (+-x,0) parts

Answer 16

lets go ahead and view this from the origin by taking out the center bits

x^2 y^2
--- - ---- = 1
1^2 2^2

when y=0, x = -1, +1



so something like this