<s>[INST] Given $x^2 + 2y^2 - 6x + 4y + 7 = 0$, find center, foci, vertex/vertices 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?
 [/INST] 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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