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").
$$A := r(1,0) \qquad B := r (\cos 2C,\sin 2C) \qquad C := r (\cos(-2B),\sin(-2B))$$
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:
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 \\
\right| = 0$$
This expands and reduces (with the help of Mathematica) to
0 &= x^2 \cos(B - C - \theta) - y^2 \cos(B - C - \theta) - 2 x y \sin(B - C - \theta) \\
&- 4 x r \cos B \cos C \cos\theta \\
&- 2 y r (\sin C \cos(B + \theta) - \cos C \sin(B - \theta))\\
&+ r^2 (2 \cos(B+C) \cos\theta + \cos(B-C+\theta)
The discriminant of this equation is
$$(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 $H$.