Question 598910
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Re-write like this:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(x\ -\ 2\)^2}{1}\ +\ \frac{(y\ -\ 3)^2}{64}\ =\ 1]



Then note that


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


is the standard form of an ellipse centered at *[tex \LARGE (h, k)], and since *[tex \LARGE a\ <\ b], semi-minor axis *[tex \LARGE a], and semi-major axis *[tex \LARGE b]


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