4
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ How would this work for a parabola? $\endgroup$
    – Ron Gordon
    Mar 25, 2013 at 14:04
  • $\begingroup$ @RonGordon If the centre exists it would satisfy this. I'm not saying that the solution will be the centre. $\endgroup$ Mar 25, 2013 at 14:05

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .