Question 618147
<font face="Times New Roman" size="+2">


The equation has both *[tex \LARGE x^2] and *[tex \LARGE y^2] terms, so it is <i>not</i> a parabola.


The equation has the same sign on the *[tex \LARGE x^2] and *[tex \LARGE y^2] terms, so it is <i>not</i> a hyperbola.


The equation has different valued coefficients on the *[tex \LARGE x^2] and *[tex \LARGE y^2] terms, so it is <i>not</i> a circle.


Hence, the equation describes an ellipse.


Complete the square on both the *[tex \LARGE x] and *[tex \LARGE y] terms, then divide both sides by the constant term remaining in the RHS so that the form looks like:


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


Since the coefficient on *[tex \LARGE x^2] is smaller than the coefficient on *[tex \LARGE y^2], the denominator below the *[tex \LARGE (x\ -\ h)^2] will be the larger denominator and since there is no *[tex \LARGE xy] term, the major axis of the ellipse will be parallel to the *[tex \LARGE x]-axis.


So, center at


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (h, k)]


Vertices at:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (h\,\pm\,a,k)]


where *[tex \LARGE a] is the square root of the denominator beneath *[tex \LARGE (x\ -\ h)^2].


Foci at: 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (h\,\pm\,c,k)]


where *[tex \LARGE c\ =\ \sqrt{a^2\,-\,b^2}], *[tex \LARGE a] is the square root of the denominator beneath *[tex \LARGE (x\ -\ h)^2] and *[tex \LARGE b] is the square root of the denominator beneath *[tex \LARGE (y\ -\ k)^2] 


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>