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I'm trying to convert the equation

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

into its standard form by choosing a new set of axes.

Yet, when I go down the conventional route, there is no xy term so $$cot2{\theta}={(a-c)/{b}}$$ doesn't work.

I've simplified it but it turns out as $$(x+2)^2 +2(y-1)^2 =1$$

So, it is a circle in standard form already?

Edward
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1 Answers1

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Note that $$(x+2)^2+2(y-1)^2=1$$ is not correct and that it is not a circle.

We have $$x^2 +2y^2 +4x-4y+4=0$$ $$\iff (x^2+4x+4)+2(y^2-2y+1)-2=0$$ $$\iff (x+2)^2+2(y-1)^2=2$$ $$\iff \frac{(x+2)^2}{\left(\sqrt 2\right)^2}+\frac{(y-1)^2}{1^2}=1$$ which is an ellipse.

mathlove
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  • Thanks. It was a slip with completing the squares. The only thing I'm confused about is why I was told to choose a ''new coordinates''? Given that it's in its standard form already. – Edward Oct 14 '14 at 19:21
  • @Yankee: You are welcome. Well, ask the person if you can:) – mathlove Oct 14 '14 at 19:25
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    @Yankee I think they are referring to the coordinates $x^{\prime}=x+2, y^{\prime}=y-1$. – user84413 Oct 14 '14 at 20:09