Thread: 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

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:
. Rearrange to:



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:



Greetings

EB

eternally grateful, i am

Thanks