1. ## completing the square

I need some help please I'm trying to rewrite the below equation into the form of a hyperbola

x^2-4x-y^2+6y-6=0

on competing the square for x^2-4x i get

x^2-4x = (x^2-4x+4)-4 = (x-2)^2-4 thats ok

But I'm getting in a terrible mess when completing the square for -y^2+6y

the final answer in the exercise book is

(x-2)^2-(y-3)^2-4+9-6=0

That is (x-2)^2-(y-3)^2=1

So it would seem that the completed square form of
-y^2+6y is (y-3)^2+9

But I just can't seem to get this I end up with a negative 9 if someone could take me through this process step by step I would be eternally grateful

2. Originally Posted by macca101
I need some help please I'm trying to rewrite the below equation into the form of a hyperbola

x^2-4x-y^2+6y-6=0

...
Hello,

you've got this:
$x^2-4x\ \ -y^2+6y-6=0$. Rearrange to:

$(x^2-4x+4) \ -(y^2-6y+9)=6+4-9$

Notice that there is a negative sign before the 2nd paranthese. Therefore the 9 which is added in the paranthese to get a complete square is actually subtracted. So you added 4 and subtracted 9 on the LHS of the equation and therefore you have to do the same on the RHS too.
You'll get:

$(x-2)^2 \ - (y-3)^2=1$

Greetings

EB

3. eternally grateful, i am

Thanks