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I read that one can calculte the center of an ellipse by calculating the partial diverative.

Let $g(x,y)=0$ be an ellipse then the center can be found by solving $\nabla g=0$

I would like to see a proof of the Lemma above (Not about general quadratic forms)

Would appreciate your help.

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    Have you tried taking an equation like $g(x,y)=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} - 1=0$ and differentiating? (After translation and rotation, that's what they all look like.) – 2017-01-13
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    $\nabla g =0$ finds the extrema of $g(x,y)$. The extrema is at the center due to the symmetry of $g(x,y) = A x^2 + B y^2 + C x y + D$ about its center. – 2017-01-13

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Consider a conic with equation $$ g(x,y)=ax^2+2bxy+cy^2+2dx+2ey+f=0 $$ Then $\nabla g=(2ax+2by+2d,2bx+2cy+2e)$ and this equals zero when $$ \begin{cases} ax+by+d=0\\ bx+cy+e=0 \end{cases} $$ This system has a unique solution if and only if $ac-b^2\ne0$ (which is the case if the conic is an ellipse or a hyperbola). If this holds and $(h,k)$ is the solution, perform the translation $$ \begin{cases} x=X+h\\ y=Y+k \end{cases} $$ to find that the equation becomes $$ aX^2+2bXY+cY^2+p=0 $$ which is the equation of a conic with its center at the origin. Thus $(h,k)$ is indeed the center of the conic. If $p=0$, the conic is degenerate.

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    If the discriminant is zero you could have a _parabola_. Parabolas are considered nondegenerate, but they do not have centers. – 2017-01-14
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    @OscarLanzi Sure. The OP was asking about ellipses, though. – 2017-01-14