Question 590121
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ -\ y^2\ +\ 4x\ +\ 4y\ -\ 4\ =\ 0]


Complete the squares:


Constant term to the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ -\ y^2\ +\ 4x\ +\ 4y\ =\ 4]


Group like variables:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ +\ 4x\ -\ y^2\ +\ 4y\ =\ 4]


Lead coefficient on *[tex \LARGE x^2] term is not 1, so factor the lead coefficient from all of the *[tex \LARGE x] terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(x^2\ +\ 2x\ \ \ )\ -\ y^2\ +\ 4y\ =\ 4]


Divide the coefficient on the 1st degree *[tex \LARGE x] term by 2, square the result, and add the result inside the parentheses.  2 divided by 2 is 1, 1 squared is 1, add 1 inside the parentheses.  Then add the lead coefficient times the amount added inside the parentheses to the RHS.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(x^2\ +\ 2x\ +\ 1)\ -\ y^2\ +\ 4y\ =\ 4\ +\ 2]


Factor the minus sign out of the *[tex \LARGE y] terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(x^2\ +\ 2x\ +\ 1)\ -\ (y^2\ -\ 4y\ \ \ ) =\ 4\ +\ 2]


Divide the coefficient on the 1st degree *[tex \LARGE y] term by 2, square the result, and add the result inside the parentheses.  4 divided by 2 is 2, 2 squared is 4, add 4 inside the parentheses.  Then add the lead coefficient, -1 here, times the amount added inside the parentheses to the RHS.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(x^2\ +\ 2x\ +\ 1)\ -\ (y^2\ -\ 4y\ +\ 4) =\ 4\ +\ 2\ -\ 4]


Factor the two perfect square trinomials and collect terms in the RHS:


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


Multiply by the reciprocal of the constant in the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(x\ +\ 1)^2}{1}\ -\ \frac{(y\ -\ 2)^2}{2} =\ 1]


Which is standard form.  However, as an aid to graphing, I like to re-write the standard form as:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\left(x\ -\ (-1)\right)^2}{1^2}\ -\ \frac{(y\ -\ 2)^2}{\sqrt{2}^2} =\ 1]


allowing you to determine the center, major axis measure (twice the square root of the denominator on the positive term, and conjugate axis measure (twice the square root of the denominator on the negative term).


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