The $x y=c^2$ problem If the points $P$, $Q$, $R$ satisfy the relation $x y=c^2$, then prove that the orthocentre formed by these $3$ points also satisfies the above relation.

 A: Forget about conic sections.
Suppose that the triangle $\triangle PQR$ is defined with the following coordinates: $P(x_1,y_1)$, $Q(x_2,y_2)$ and $R(x_3,y_3)$. Obviously:
$$x_iy_i=c^2$$
The slope of line $PQ$ is:
$$k=\frac{y_2-y_1}{x_2-x_1}=\frac{\frac{c^2}{x_2}-\frac{c^2}{x_1}}{x_2-x_1}=-\frac{c^2}{x_1x_2}$$
The slope of line perpendicular to $PQ$ is:
$$k'=-\frac{1}{k}=\frac{x_1x_2}{c^2}$$
Let's find the equation of the line passing through $R$ perpendicular to $PQ$. The orthocenter $(x_H,y_H)$ must be on that line:
$$y_H-y_3=k'(x_H-x_3)$$
$$y_H-\frac{c^2}{x_3}=\frac{x_1x_2}{c^2}(x_H-x_3)\tag{1}$$
You can repeat the same proces for triangle side $QR$ by finding a line perpendicular to it passing through point $P$. The orthocenter $(x_H,y_H)$ must be on that line too:
$$y_H-\frac{c^2}{x_1}=\frac{x_2x_3}{c^2}(x_H-x_1)\tag{2}$$
You have two linear equations, (1) and (2) with two unknowns $(x_H,y_H)$. Solving them is trivial and the final result is:
$$x_H=-\frac{c^4}{x_1x_2x_3}, \quad y_H=-\frac{x_1x_2x_3}{c^2}$$
Obviously $x_Hy_H=c^2$ so the orthocenter lies on the hyperbola too.
A: Hint:  
Assume $P,Q,R$ lie on a Rectangular Hyperbola $xy=c^2$.  
$P,Q,R$ will form an Obtuse-Angled Triangle whose Orthocentre will also lie on the same Hyperbola   (outside triangle PQR), so that locus of all $4$
 points ($P,Q,R$ and $O$) trace $xy=c^2$
