Find centers of chords of quadratic form

Question

Prove that the centers of chords of quadratic form $$f(x,y) = A x^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$ that are parellels to a line $\frac{x}{\xi}=\frac{y}{\eta}$ lies on $$ \left( \frac{\partial f}{\partial x} \right) \xi + \left( \frac{\partial f}{\partial y} \right) \eta = 0$$

Answer 1

Setting apart the exceptional case $\eta=0$ (which can easily be treated separately), we can set

$$a=\dfrac{\xi}{\eta}.$$

Therefore, we have a reference straight line $y=ax$.

All straight line parallel to it have equations

$$y=ax+b \ (1a) \ \ \ \ \iff \ \ \ \ x=\dfrac{y-b}{a} \ (1b)\tag{1a/1b}$$

The intersections of conic curve

$$f(x,y) = A x^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \tag{2}$$

are given

• for the abscissas, by the solutions of quadratic equation in variable $x$ obtained by substituting $f(x,ax+b)=0$ (using (1a)). But in fact, we do not need to obtain the abscissas themselves, but their half sum, which is easy on a quadratic $ux^2+vx+w=0$ (it is equal to $-\dfrac{v}{2u}$). All computations done, we obtain, for the current abscissas of midpoints :

$$x_m=\dfrac{-(B+2Ca)b - (D+Ea)}{2(A+Ba+Ca^2)}\tag{3}$$

• for the ordinates, same work, but with substitution (1b) : $f(\dfrac{y-b}{a},y)=0$ giving this time a quadratic in $y$, out of which one obtains the current ordinate of the midpoint :

$$y_m=\dfrac{ (2A+Ba)b - (D + Ea)a}{2(A+Ba+Ca^2)}\tag{4}$$

One can already see that

$$x_m=pb+q, \ \ \ y_m=rb+s\tag{5}$$

for certain constants $p,q,r,s$, proving that the locus is (part of) a line. (Remember that $a$ is fixed and $b$ is variable).

From there, it shouldn't be difficult to prove that (once we have divided by $\eta$ the given relationship with partial derivatives) that (5) verifies :

$$(2Ax_m+By_m+D)+(Bx_m+2Cy_m+E)a=0$$

(indeed the partial derivatives have to be computed in $(x_m,y_m)$).

Remark : as I already said, there is a whole classical theory behind this, the theory of conjugate diameters (dating back to Appolonius). See for example.

Important edit : Some hours later, I realized that the calculations above can be enormously shortened by considering that the property of the midpoints to be established is invariant by translation and rotation. Therefore, we can work with reduced equations, distinguishing :

$$x^2/a^2+y^2/b^2=1 \ \text{(ellipse)}, \ \ x^2/a^2-y^2/b^2=1 \ \text{(hyperbola)}, \ \ y^2-2px=0 \ \text{(parabola)}.$$