1
$\begingroup$

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?

  • 1
    What asymptote? This is an ellipse. And I get a different "standard form."2012-06-10
  • 0
    @AndréNicolas perhaps the OP means the directrix?2012-06-10
  • 0
    Why do you divide both $x^2$ and $y^2$ by four in your transformed equation?2012-06-10
  • 0
    For future reference, you may be able to get quicker answers from [Wolfram Alpha](http://www.wolframalpha.com/input/?i=x%5E2%2B2y%5E2%E2%88%926x%2B4y%2B7%3D0)2012-06-10
  • 0
    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.)2012-06-10
  • 0
    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

1 Answers 1

2

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.

  • 0
    Is the focal axis parallel to the x-axis?2012-06-10
  • 0
    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.)2012-06-10
  • 0
    Hmm, howcome you said the verticies have to change? Arnt they at (1,-1) and (5,-1)?2012-06-10
  • 0
    Ah! Right you are. I'll fix it. (Bonehead mistake. Forgot to take the square root.)2012-06-10
  • 0
    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