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Say we have a conic with equation $f(x,y)=c$.

My teacher says that it's centre satisfies the equations :

$f_x(x,y)=f_y(x,y)=0$ (If it has a centre).

She didn't give any explanation. I thought this was because if we have centre $(x_0,y_0)$, then $f(x_0+x,y_0+y) =f(x_0-x,y_0-y)$(As it's a centre, the point diametrically opposite is on the curve). Differentiating, we get $f_x(x_0+x,y_0+y) =-f_x(x_0-x,y_0-y)$ and setting $x=y=0$, we get the result. Is this correct? I realize that I have assumed the entire family of curves $f(x,y)=c$ have the same centre,how do I avoid that? Is this valid for other curves?

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How would this work for a parabola? – Ron Gordon Mar 25 '13 at 14:04
@RonGordon If the centre exists it would satisfy this. I'm not saying that the solution will be the centre. – Ishan Banerjee Mar 25 '13 at 14:05
up vote 2 down vote accepted

You did not assume that every curve in the family $f(x,y) = c$ has the same center. You showed that for any curve in that family with a particular center $(x_0,y_0)$, then the center satisfies that differential relation.

Fortunately, every curve with a center has a center expressible as a coordinate pair $(x_0,y_0)$, so your proof holds just fine. Further, your proof readily generalizes to curves in higher dimensions with a center.

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