Proof of midpoint theorem in hyperbolas/conics Please help me find a proof for the midpoint theorem in hyperbolas (or conics in general).
All midpoints of parallel chords in a hyperbola/conic are located on a common line.
Motivation: I study mathematics for teaching and we are doing some basic geometry. We covered hyperbolas on a basic level, i.e. tangents and intersections with hyperbolas. We also discussed the midpoint theorem in hyperbolas, but unfortunately I haven't found a proof that would be comprehensible to high school students. My first idea is to find two midpoints of parallel chords and show that they are located on the same line.


 A: 
Fig. 1: If the common slope of the secant lines is $m$, the slope of the line containing all midpoints is $\tfrac{1}{m}$.
As you ask the question in the framework of teachers' training, I would advise to take the simplest hyperbola, i.e., the equilateral hyperbola, and restrict the study to the upper branch with equation
$$y=\sqrt{1+x^2}\tag{1}$$
Consider parallel lines having equation
$$y=mx+p\tag{2}$$
($m$ fixed, variable $p$). Intersection points $P_k(x_k,y_k)$ $(k=1,2)$ (when they exist) of the hyperbola branch with these lines have their abscissas $x_k$ solutions of equation (I jump over details that are evident to us ; some recalls are probably necessary here for your pupils):
$$\sqrt{1+x^2}=mx+p$$
which are the solutions of quadratic equation :
$$\underbrace{(1-m^2)}_a x^2+\underbrace{(-2mp)}_b x+\underbrace{(1-p^2)}_c=0$$
The half-sum of the roots is classicaly given by
$$x_h=-\tfrac{b}{a}=\underbrace{\tfrac{m}{1-m^2}}_k p\tag{3}$$
(no need to compute explicitly the roots), and this $x_h$ is the abscissa of the midpoint. Plugging expression (3) into (2) gives
the ordinate of the midpoint :
$$y_h=m x_h + p = \tfrac{m^2}{1-m^2} p + p = \tfrac{1}{1-m^2} p \tag{4}$$
By elimination of $p$ between equations (3) and (4), we get a linear correspondence
$$x_h=m y_h$$
Otherwise said, the locus belongs to the straight line $y=\frac{1}{m}x.$
Why belongs instead of is ? One can verify graphically that this line isn't exactly the locus : one has to eliminate a segment of line. The students can be asked : has algebra "lied" ? Answer : evidently no : look at the quadratic equation : has it alwyas solutions ? Etc.
Essential remark (done by Intelligenci Pauca): Had we taken, instead of (1) the equilateral hyperbola under the form $y=\frac{1}{x}$, we could have simpler computations with the remarkable fact that lines with common equation (2) with slope $m$ generate a midpoints' locus with slope $-m$.
This correspondence :
$$m  \ \ \leftrightarrow \ \ -m$$
mirroring the correspondence :
$$m  \ \ \leftrightarrow \ \ \frac{1}{m}$$
obtained above.
Remark 1 : Once this is done for this particular case of hyperbola, one can extend the result to any hyperbola by explaining to students that is the image of an equilateral hyperbola by an affine transformation, knowing (is it the case or do they have to admit it ?) that affine transformations preserve parallelism and midpoints.
Remark 2 : The "theoretical concept" behind this property is captured into the expression "conjugate diameters".
A: The equation of a hyperbola or an ellipse centered at the origin is given by
$r^T Q r = 1$
where $ Q $ is a symmetric $2 \times 2 $ matrix.  Now consider a line passing through $p_0$ and having a direction vector $ v_1$, then it is given parametrically by,
$r = p_0 + t v_1$
Plug this into the equation of the hyperbola (or ellipse) you get
$ (p_0 + t v_1)^T Q (p_0 + t v_1) = 1 $
Expanding,
$t^2 (v_1^T Q v_1) + 2 t p_0^T Q v_1 + p_0^T Q p_0 - 1 = 0 $
If the line intersects the hyperbola (ellipse), the roots of this quadratic equation are $ t_1, t_2 $ that correspond to two points on the line
$r_1 = p_0 + t_1 v_1 , \ r_2 = p_0 + t_2 v_1 $
The midpoint of $r_1, r_2$ is given by
$ r^* = p_0 + \left( \dfrac{t_1 + t_2}{2} \right) v_1 $
And thus corresponds to a value of $t$ that is the midpoint of $(t_1, t_2)$ , the two roots of the quadratic equation, and we know that this is given by $\dfrac{-B}{2 A} $ , i.e.
$t^* =   - \dfrac{p_0^T Q v_1}{ v_1^T Q v_1 } $
Hence, explicitly, the midpoint is given by,
$r^* = p_0 - v_1 \dfrac{v_1^T Q p_0}{  v_1^T Q v_1 } = \left( I - \dfrac{v_1 v_1^T Q}{  v_1^T Q v_1} \right) p_0 = A p_0 $
Now let $v_2$ be perpendicular to $v_1$, then we can write
$p_0 =  t v_1 +  s v_2 $
It follows that the midpoint with this $p_0$ is given by
$r^* = A ( t v_1 + s v_2) =  t A v_1 + s A v_2 $
it is easy to check that $A v_1 = 0$ , therefore,
$r^* = s A v_2$
which is an equation of a straight line passing through the origin (which is the center of the hyperbola (or ellipse) ) and having a direction vector $A v_2$.  This completes the proof.
For a parabola, similar arguments applied the parabola model lead to the same conclusion.
For completeness, I will include the parabola case.
The parabola algebraic equation is
$(r - V)^T Q (r - V) + b^T (r - V) = 0$
where $V$ is the vertex of the parabola, and $Q = R D R^T$ with $D = \text{ diag}\{ a, 0 \}$ and $R$ a rotation matrix, and $b^T = b_0^T R$ , where $b_0= [0, -1]^T$, so $b = R^T b_0$.
Again a straight line is given by
$r = p_0 + t v_1$
Plug this in, you get
$(p_0 - V + t v_1) Q (p_0 - V + t v_1) + b^T (p_0 - V + t v_1) = 0 $
which is a quadratic equation in $t$, and again the midpoint of the two intersections corresponds to the midpoint of $t_1, t_2$ the two roots
of this quadratic, which occurs at
$-\dfrac{B}{2A} = - \dfrac{ 2(p_0 - V)^T Q v_1 + b^T v_1 }{  2 v_1^T Q v_1} = - \dfrac{ (2(p_0 - V)^T Q + b^T ) v_1 }{ 2 v_1^T Q v_1}$
Therefore, the midpoint is given by,
$r^* = p_0 - v_1 \dfrac{ \left( 2(p_0 -V)^T Q + b^T \right) v_1 } { 2 v_1^T Q v_1 }$
After simple manipulation, this reduces to,
$r^*= v_1 \dfrac{2 V^T Q v_1- b^T v_1  }{ 2 v_1^T Q v_1 } + \left(I - \dfrac{ v_1 v_1^T Q }{v_1^T Q v_1} \right) p_0 $
written compactly,
$r^* = p_1 + A p_0 $
where
$p_1 =v_1 \dfrac{  2 V^T Q v_1 - b^T v_1  }{ 2 v_1^T Q v_1 } $
and
$ A = I - \dfrac{ v_1 v_1^T Q }{v_1^T Q v_1} $
Using $v_2$, a perpendicular vector to $v_1$ we can express $p_0$ as follows
$p_0 = t v_1 + s v_2 $
Using the fact that $A v_1 = 0 $, the midpoint is given by
$r^* = p_1 + s A v_2 $
which is a straight line passing through $p_1$ and having a direction vector $A v_2$.
