Question 372291
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You don't have any first degree terms, so completing the square doesn't buy you anything.  You have standard form when it looks like:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x^2}{a^2}\ +\ \frac{y^2}{b^2}\ =\ 1]


So divide by 36:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x^2}{9}\ +\ \frac{y^2}{4}\ =\ 1]


When you have first degree terms, you complete the square on each of the variables separately giving you some thing like:


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


Which centers your ellipse at *[tex \Large (h, k)] rather than the origin as in your example.


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