Question 566563
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Your -8 issue is a self-inflicted wound.  Properly done, -8 never appears anywhere in this problem.  Also, take care how you type your expressions in the future:  X and x are NOT the same thing, so technically you have a three variable equation.  Just sayin'


First complete the squares:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 10x\ +\ 37\ +\ 4y^2\ +\ 16y\ =\ 0]


Constants on the right!


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 10x\ +\ \ \ \ +\ 4y^2\ +\ 16y\ =\ -37]


Divide the 1st order *[tex \Large x] term coefficient by 2, square the result, add the new constant to both sides:  -10 divided by 2 is -5, -5 squared is 25.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 10x\ +\ 25\ +\ 4y^2\ +\ 16y\ =\ -37\ +\ 25] 


Factor a 4 out of both *[tex \Large y] terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 10x\ +\ 25\ +\ 4\left(y^2\ +\ 4y\ \ \ \right)\ =\ -37\ +\ 25]


4 divided by 2 is 2. 2 squared is 4.  4 inside the parentheses, 16 outside.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 10x\ +\ 25\ +\ 4\left(y^2\ +\ 4y\ +\ 4 \right)\ =\ -37\ +\ 25\ +\ 16]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 5)^2\ +\ 4(y\ +\ 2)^2\ =\ 4]


Divide through by the RHS constant so that the RHS becomes 1:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(x\ -\ 5)^2}{4}\ +\ \frac{(y\ +\ 2)^2}{1}\ =\ 1]


Compare to the equation of an ellipse centered at *[tex \Large (h,k)], axes parallel to the coordinate axes, major axis measuring *[tex \Large 2a], minor axis measuring *[tex \Large 2b], and foci at a distance of *[tex \Large \sqrt{a^2\ -\ b^2}] from the center on the major axis.


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


So your center is at *[tex \Large (5,-2)].  Foci at *[tex \Large (5\ +\ \sqrt{3},-2)] and *[tex \Large (5\ -\ \sqrt{3},-2)].  Major axis measures 4.  Minor axis measures 2.


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