Prove that a conic section through the vertices of a triangle and its orthocentre is a rectangular hyperbola. I can see that a rectangular hyperbola through the vertices of a triangle $ABC$ passes through the orthocentre $H$. But how do you prove the converse that a circumhyperbola through $H$ is rectangular?
Any proofs or hints would be appreciated! Thanks
 A: I'll prove this result:

Given $\triangle ABC$ with orthocenter $H$, let $P$ be any distinct point on the triangle's circumcircle. These five points determine a rectangular hyperbola (whose center happens to be on the triangle's "nine-point circle").


Coordinatize with
$$A := r(1,0) \qquad B := r (\cos 2C,\sin 2C) \qquad C := r (\cos(-2B),\sin(-2B))$$
where $r$ is the circumradius (and $A$, $B$, $C$ also serve to name the angles at each vertex), so that
$$H = r(1 - 2 \cos A \cos(B - C), 2 \cos A \sin(B - C))$$
We can take $P := r (\cos\theta, \sin\theta)$.
The unique conic through these points has this equation:
$$\left|
\begin{array}{cccccc}
x^2 & x y & y^2 & x & y & 1 \\
A_x^2 & A_x A_y & A_y^2 & A_x & A_y & 1 \\
B_x^2 & B_x B_y & B_y^2 & B_x & B_y & 1 \\
C_x^2 & C_x C_y & C_y^2 & C_x & C_y & 1 \\
H_x^2 & H_x H_y & H_y^2 & H_x & H_y & 1 \\
P_x^2 & P_x P_y & P_y^2 & P_x & P_y & 1 \\
\end{array}
\right| = 0$$
This expands and reduces (with the help of Mathematica) to
$$\begin{align}
0 &= x^2 \cos(B - C - \theta) - y^2 \cos(B - C - \theta) - 2 x y \sin(B - C - \theta) \\ 
&- 4 x r \cos B \cos C \cos\theta \\
&- 2 y r (\sin C \cos(B + \theta) - \cos C \sin(B - \theta))\\
&+   r^2 (2 \cos(B+C) \cos\theta + \cos(B-C+\theta)
) 
\end{align}$$
The discriminant of this equation is
$$(2\sin(\cdots))^2 - 4(\cos(\cdots))(-\cos(\cdots)) = 4(\sin^2(\cdots)+\cos^2(\cdots)) = 4$$
which, being positive, indicates that the represented conic is a hyperbola. That the sum of $x^2$ and $y^2$ coefficients vanishes indicates that the conic is specifically a rectangular hyperbola. (See Wikipedia's "Discriminant" subsection of its "Conic section" entry.) $\square$
Showing that the center of the conic lies on the nine-point circle is left as an exercise to the reader. (The Penguin entry referenced by @Jan-MagnusØkland's comment to the question affirms this fact.) I'll note, however, that this property was leveraged in the conception of this proof: It is known that the nine-point circle is the dilation, with scale factor $1/2$, of the circumcircle with respect to the orthocenter; consequently, for any center, $K$, chosen on the nine-point circle, the reflection of $H$ in $K$ would lie on the circumcircle and on the corresponding hyperbola. We took our $P$ to be that reflection of $H$. 
