Question

I was asked to find the standard form of the equation of this hyperbola 9x^2-y^2 + 54x + 10y + 55 = 0 but as result I got 9(x+3)^2 - (y+5)^2 = 1 and I don't know what to do with the 9...

Answer 1

you completed your squares?

Answer 2

9x^2 -y^2 + 54x + 10y + 55 = 0

9x^2 + 54x -y^2 + 10y + 55 = 0

9(x^2 + 6x) -(y^2 -10y) + 55 = 0

9(x^2 + 6x+ 9-9) -(y^2 -10y +25-25) + 55 = 0

9((x+3)^2-9) -((y-5)^2-25) + 55 = 0

9(x+3)^2 -9(9) -(y-5)^2 -(-25) + 55 = 0

9(x+3)^2 -81 -(y-5)^2 +25 + 55 = 0

9(x+3)^2 -81 -(y-5)^2 +80 = 0

9(x+3)^2 -(y-5)^2 = 1

Answer 3

yeah, thats all there is to it

Answer 4

Thanks, I think that in order to have the equation in standard form I need to
(x+3)^2 - (y+5)^2/9 = 1/9
(x+3)^2/1/2 - (y+5)^2 = 1

Answer 5

Sorry (x+3)^2/1/9 *

Answer 6

no, you have it fine as is

Answer 7

its equal to 1, you cant divide off the 9 and still have it =1

and having

(x-__)^2
-----
(1/9)

just isnt proper that im aware of

Answer 8

if we had something like =6, we would have divided off the 6 to get it into standard form

9/6 = 3/2 and 1/6 is well 1/6 other than that