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So the equation is:

$ x^2 + 2y^2 - 6x + 4y + 7 = 0 $

Find the coordinates of the center, the foci, and the vertex or vertices.

What I did was put the equation in the form: $ \frac{(x-3)^2}{4}+ \frac{(y+1)^2}{4} = 1 $

Now based on that, I said the center is at $(3,-1)$, the foci is at ~+- 2.45 (since $c = \sqrt {a^2 + b^2}$ ). so the coordinates of that are $(3+2.45,-1)$ and $(3-2.45,-1)$ and the vertices are $(1,-1)$ and $(5,-1)$. I also went ahead and found the asymptote, which is just done by setting the equation to $0$, correct?

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    If what you wrote really was the correct formula then you'd in fact have a *circle*, thus making all the questions about foci, radiuses, foci, etc. pretty easy to answer but also pretty boring.2012-06-11

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The equation should be $\frac{(x-3)^2}{4}+\frac{(y+1)^2}{2}=1.$ You've correctly identified the center and vertices. The focal length should be $\sqrt{a^2-b^2}$, not $\sqrt{a^2+b^2}$. Ellipses don't have asymptotes, you're thinking of hyperbolae.

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    Awesome. Yes I made a typo in my original question, I did get the same solution as you. But I got the foci wrong because I did + instead of -... thanks alot!2012-06-10