TITLE: Why does partial differentiation give centre of a conic?
QUESTION [16 upvotes]: Definition of centre of conics states that, "The point which bisects every chord of the conic passing through it is called centre of the conic"
General equation of second degree is $S:ax^2+2hxy+by^2+2gx+2fy+c=0$ If we partially differentiate $S$ with respect to $x$ and $y$, and then set both partial derivatives to $0$  . Solving two equations give centre of conics. Could someone explain simple logic behind this approach i.e. why we get centre using above methodology?

REPLY [6 votes]: Let $Q(x,y)=ax^2+bxy+cy^2+dx+ey+f$ be a polynomial in $x$ and $y$. The conic is, of course, the set of its zeros. By studying the symmetries of the figure, we find that the center, if it exists is a point $(P_x,P_y)$ such that $Q(x+P_x,y+P_y)=a'x^2+b'xy+c'y^2+f'$; i.e. it does not have any terms of degree 1. 
For a polynomial in one variable, $P(x)=a_0+a_1x+\cdots+ a_nx^n$, and a constant $a$, we can find the coefficients of $P(x+a)$ using Taylor's theorem: $P(x+a)=P(a)+P'(a)x+P''\frac{x^2}{2!}+\cdots+ P^{(n)}(a)\frac{x^n}{n!}$. Analogously, using the generalization of Taylor's theorem to more than one variable, we find that 
$$Q(x+P_x,y+P_y)=Q(P_x,P_y)+\frac{\partial Q}{\partial x}(P_x,P_y)x+\frac{\partial Q}{\partial y}(P_x,P_y)y+\frac{\partial^2 Q}{\partial x^2}(P_x,P_y)\frac{x^2}{2}+\frac{\partial^2 Q}{\partial y^2}(P_x,P_y)\frac{y^2}{2}+\frac{\partial^2 Q}{\partial x \partial y}(P_x,P_y)xy.$$
But we want the coefficients of the terms of degree 1 to be zero; so, if such a point exists, then $\frac{\partial Q}{\partial x}(P_x,P_y)=\frac{\partial Q}{\partial y}(P_x,P_y)=0$.