Question 629704
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Keeping the signs straight when completing the square on the general quadratic for a hyperbola can be a little tricky.  Watch carefully:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ y^2\ +\ 2x\ +\ 6y\ -\ 9\ =\ 0]


Move the constant term to the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ y^2\ +\ 2x\ +\ 6y\ =\ 9]


Rearrange:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 2x\ -\ y^2\ +\ 6y\ =\ 9]


Since you are working a hyperbola, group the "negative variable" terms with parentheses:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 2x\ -\ (y^2\ -\ 6y\ )\ =\ 9]


Note the sign change on the first degree *[tex \LARGE y] term because of the distributive law.


Now complete the square on both variables:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 2x\ +\ 1\ -\ (y^2\ -\ 6y\ +\ 9)\ =\ 9\ +\ 1\ -\ 9]


Notice that because you are adding a +9 INSIDE of the parentheses on the left, you are actually adding a -9 to the LHS.  So you need a -9 in the RHS to make up for it.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 1)^2\ -\ (y\ -\ 3)^2\ =\ 1].


so center at *[tex \LARGE (-1, 3)] just like you had it, but *[tex \LARGE a\ =\ 1], *[tex \LARGE b\ =\ 1], and therefore *[tex \LARGE c\ =\ \sqrt{2}]


I think you can take it from here, can you not?


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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