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Find the centre of the ellipse $$14x^2-4xy+11y^2-44x-58y+71=0$$

My attempt : I know from the generic conic equation $ax^2+by^2+2hxy+2gx+2fy+c=0$ that the condition for an ellipse is $h^2<ab$. But I am unaware of the procedure to find the centre from the generic equation. Pls help.

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Let $f(x,y)=14x^2-4xy+11y^2-44x-58y+71=0$ and solve the system \begin{cases} \dfrac{\partial f}{\partial x}=0,\\ \dfrac{\partial f}{\partial y}=0. \end{cases} then \begin{cases} 28x-4y-44=0,\\ -4x+22y-58=0. \end{cases} so the center is $(2,3)$.

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