standard form of hyperbola

[daemon]

Hi,

How would I convert this equation to the standard form of a hyperbola by completing the square on x and y?

x^2 - y^2 + 6x + 2y + 7 = 0

 

[pka]

What have you done?
Where do you have problems?
Please show some effort on your part!

 

[tkhunny]

Personally, I would complete the square on x and y.

Have you REALLY no idea how to do that?

 

[Unco]

Hi,

How would I convert this equation to the standard form of a hyperbola by completing the square on x and y?

x^2 - y^2 + 6x + 2y + 7 = 0

Group x and y terms together:

(x^2 + 6x) - (y^2 - 2y) = -7

Complete the square in each set of parentheses.

x^2 + 6x = (x + 3)^2 - 9

etc.

 

[daemon]

I came up with (x+3)^2 - (y-1)^2 = 1

The thing I was confused about was that from what I've read the standard form of hyperbola was either

(x-h)^2 / (a^2) - (y-k)^2 / (b^2) = 1

or

(y-k)^2 / (b^2) - (x-h)^2 / (a^2) = 1

My answer doesn't contain a denominator like the standard forms of hyperbolas I have seen, that was what I was unsure about.

 

[Unco]

Your answer is absolutely correct. There's nothing wrong with a=1 and b=1.