# 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?

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What asymptote? This is an ellipse. And I get a different "standard form." – André Nicolas Jun 10 '12 at 20:11
@AndréNicolas perhaps the OP means the directrix? – Shaktal Jun 10 '12 at 20:13
Why do you divide both $x^2$ and $y^2$ by four in your transformed equation? – krlmlr Jun 10 '12 at 20:14
For future reference, you may be able to get quicker answers from Wolfram Alpha – Daniel Littlewood Jun 10 '12 at 20:15
Note the spelling of the plural of "vertex". It is "vertices", just as the plural of "matrix" is "matrices", of "axis" is "axes", of "basis" is "bases", of "thesis" is "theses", etc. (I corrected it in several places in the posting.) – Michael Hardy Jun 10 '12 at 20:29

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.
Since $4>2$, yes. If it had been (for example) $\dfrac{(x-3)^2}{2}+\dfrac{(y+1)^2}{4}=1$, then the focal axis would have been parallel to the $y$-axis, instead. (Recall that the focal axis is the major axis.) – Cameron Buie Jun 10 '12 at 20:55