Question 372311
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(x\ -\ h)^2}{a^2} \pm \frac{(y\ -\ k)^2}{b^2}\ =\ 1]


If you have a plus sign between the LHS terms, and...


If *[tex \LARGE a\ =\ b], you have a circle center at *[tex \LARGE (h,k)], radius *[tex \LARGE a], but if *[tex \LARGE a\ \neq\ b] you have an ellipse, center at *[tex \LARGE (h,k)], semi-major axis *[tex \LARGE a] and semi-minor axis *[tex \LARGE b] (unless *[tex \LARGE b > a] and it is the other way around).


If you have a minus sign, then you have a hyperbola, centered at *[tex \LARGE (h, k)]


If only one of the variables is squared, then you have a parabola.


Only hyperbolas have asymptotes.


For your example, *[tex \LARGE a\ >\ b], *[tex \LARGE h\ =\ 0], and *[tex \LARGE k\ =\ 0].  So you have an ellipse, centered at the origin, semi-major axis of 3, and semi-minor axis of 2.  The vertices are at *[tex \LARGE (3,0)] and *[tex \LARGE (-3,0)], the end points of the semi-minor axis are at *[tex \LARGE (0,2)] and *[tex \LARGE (0,-2)].  Calculate *[tex \LARGE c\ =\ \sqrt{a^2\ -\ b^2}\ =\ \sqrt{9\ -\ 4}\ =\ \sqrt{5}].  Then the foci are at *[tex \LARGE (\sqrt{5},0)] and *[tex \LARGE (-\sqrt{5},0)].  Ellipses do not have asymptotes.


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