Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.