# Ellipse with center in origin

The purpose is to fit data to a ellipse which center is the origin $(x_0=0,y_0=0)$.

I found the general quadratic curve: $$ax^2+2bxy+cy^2+2dx+2fy+g=0$$ Reference: http://mathworld.wolfram.com/Ellipse.html (eq. 15)

To get to this, I use the simple ellipse equation: $$\frac{x^2}{a_e^2}+\frac{y^2}{b_e^2}=1$$ With the substitutions $x\rightarrow x^{\prime}=x-x_0$, $y\rightarrow y^{\prime}=y-y_0$ and the rotation $x^{\prime}\rightarrow x^{\prime \prime}=\cos\phi x^{\prime}+\sin\phi y^{\prime}$, $y^{\prime}\rightarrow y^{\prime \prime}=-\sin\phi x^{\prime}+\cos\phi y^{\prime}$, I can express the parameters above in terms of $x_0$, $y_0$, $a_e$,$b_e$ and $\phi$: $$a=\frac{\cos^2\phi}{a_e^2}+\frac{\sin^2\phi}{b_e^2}$$ $$2b=2\sin\phi\cos\phi\left(\frac{1}{a_e^2}-\frac{1}{b_e^2}\right)$$ $$c=\frac{\sin^2\phi}{a_e^2}+\frac{\cos^2\phi}{b_e^2}$$ $$2d=-2x_0\left(\frac{\cos^2\phi}{a_e^2}+\frac{\sin^2\phi}{b_e^2}\right)-2\sin\phi\cos\phi y_0\left(\frac{1}{a_e^2}-\frac{1}{b_e^2}\right)$$ $$2f=-2y_0\left(\frac{\sin^2\phi}{a_e^2}+\frac{\cos^2\phi}{b_e^2}\right)-2\sin\phi\cos\phi x_0\left(\frac{1}{a_e^2}-\frac{1}{b_e^2}\right)$$ $$g=\left(\cos\phi x_0+\sin\phi y_0\right)^2\left(\frac{1}{a_e^2}+\frac{1}{b_e^2}\right)-1$$

Does someone have a reference to check this relations?

The general quadratic curve has 6 parameters $(a,b,c,d,f,g)$, which are expressed by 5 properties of the ellipse. There is a further limitation in order to have an ellipse: $$ac-b^2>0$$ Has anyone an idea how to understand this inequality?

Since I want to fit to a ellipse with $(x_0=0,y_0=0)$, I now get: $$d=f=0$$ $$g=-1$$ So, now my main question is: Why do I get a constraint on 3 parameters only by placing the ellipse in the origin?

If you're wondering why you only imposed two conditions ($x_0=0$, $y_0=0$) and somehow obtained three, here's why. Your original equation has a redundant degree of freedom: replace $a,b,\ldots,g$ with say $2a,2b,\ldots,2g$ and the ellipse doesn't change. The second equation does not have this redundancy. So when you compared the two equations, you implicitly lost a degree of freedom: you now had six variables defined in terms of five. The only thing that happened when you plugged in $x_0=0$, $y_0=0$ is that this lost degree of freedom became explicit.
A simpler example to help the intuition: Suppose you have $x=a, y=b, z=ab$. Add a single condition $b=0$ and you magically get not one but two equations, $y=0$ and $z=0$.