Hyperbola

• Jun 4th 2010, 06:48 PM
Honorable24
Hyperbola
-x^2 + y^2 -2x -12y + 31 =0

(y^2-12y +36) - (x^2 -2x +1) = -31 +36 -1

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

Did I do this right?
• Jun 4th 2010, 06:55 PM
Prove It
Quote:

Originally Posted by Honorable24
-x^2 + y^2 -2x -12y + 31 =0

(y^2-12y +36) - (x^2 -2x +1) = -31 +36 -1

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

Did I do this right?

You have made a mistake...

$\displaystyle -x^2 + y^2 - 2x - 12y + 31 = 0$

$\displaystyle y^2 - 12y - (x^2 + 2x) + 31 = 0$

$\displaystyle y^2 - 12y + \left(-6\right)^2 - [x^2 + 2x + 1^2] + 31 - \left(-6\right)^2 + 1^2 + 31 = 0$

$\displaystyle (y - 6)^2 - (x + 1)^2 - 36 + 1 + 31 = 0$

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

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

$\displaystyle \frac{(y - 6)^2}{4} - \frac{(x + 1)^2}{4} = 1$.
• Jun 4th 2010, 07:03 PM
sa-ri-ga-ma
http://www.mathhelpforum.com/math-he...8e60396f-1.gif

It should be

$\displaystyle (y-6)^2 - (x+1)^2 - 4 = 0$
• Jun 4th 2010, 07:09 PM
Honorable24
So I was right except the (x-1)^2 i had?
• Jun 4th 2010, 07:09 PM
Prove It
Quote:

Originally Posted by sa-ri-ga-ma
http://www.mathhelpforum.com/math-he...8e60396f-1.gif

It should be

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

Thanks, stupid typos hahaha.