The normals at three points $P$, $Q$, $R$ on a rectangular hyperbola $xy = c^2$ intersect at a point on the curve. Prove that the centre of the hyperbola is the centroid of the triangle $PQR$.
I copied the the question word by word. Let assume that the points are (c$ t_1$,$\frac{c}{ t_1}$) , (c$ t_2$,$\frac{c}{ t_2}$) & (c$ t_3$,$\frac{c}{t_3}$), even if I draw the normal, at all the three point how do I intersect three lines at one point.
On
Let normal to the hyperbola at $(ct,\frac{c}{ t})$ is $t^{2}x-t^{3}c=y- \frac{c}{t}$
Let the common point where all the normal intersect is $(ct',\frac{c}{ t'})$
Plugging this in the equation we get the form as
$(t^{3} + \frac{1}{ t'})(t'-t)=0$
$t^{3} + \frac{1}{ t'}=0$
Coefficient of $t^2 =0$ & coefficient of $t =0$ viz.
$ \ t_1+ \ t_2+ \ t_3 =0$
&
$ \ t_1\ t_2+ \ t_2\ t_3+ \ t_1\ t_3 =0$.
Hence centriod lies on origin.
On
A triangle with vertices on the hyperbola cannot have the origin as centroid
proof
Let $y=\dfrac{c^2}{x}$ be the equation of the hyperbola and $A\left(a,\dfrac{c^2}{a}\right);\;B\left(b,\dfrac{c^2}{b}\right)$ two points on the hyperbola
The third vertex $C$ has coordinates $C\left(x,\dfrac{c^2}{x}\right)$
The relation $a+b+x=0\to x=-a-b$ must held in order to have origin $O(0,0)$ centroid of the triangle $ABC$. Therefore the point $C$ has actual coordinates $C\left(-a-b,\dfrac{c^2}{-a-b}\right)$
Consider now the $y-$coordinates of the vertices of the triangle $ABC$
They must satisfy the relation $\dfrac{y_A+y_B+y_C}{3}=0$ that is $y_A+y_B+y_C=0$
$\dfrac{c^2}{a}+\dfrac{c^2}{b}+\dfrac{c^2}{-a-b}=\dfrac{c^2 \left(a^2+a b+b^2\right)}{a b (a+b)}$
$a^2+ab+b^2\ne 0$ for any $a,b\in\mathbb{R}$ and this concludes the proof.
remark
As no triangle having vertices on the hyperbola can have centroid in the origin to a greater extent the proposition in the Original Post is false since triangle $PQR$ cannot exist
Points $P$, $Q$, $R$ cannot exist as stated. At least two among them must lie on the same branch of the given hyperbola, so we may assume for instance that $P$ and $Q$ are given by $$ P=(x_1,c^2/x_1),\quad Q=(x_2,c^2/x_2),\quad \hbox{with $x_1>0$ and $x_2>0$}. $$ The intersection point of the normals at $P$ and $Q$ is $$ N=\left(\frac{c^4+{x_1}^3 {x_2}+{x_1}^2 {x_2}^2+{x_1} {x_2}^3} {{x_1}{x_2}(x_1+x_2)}, \frac{c^4 {x_1}^2+c^4 {x_1}{x_2}+c^4 {x_2}^2+{x_1}^3{x_2}^3} {c^2 {x_1} {x_2}(x_1+x_2)}\right) $$ and $$ x_Ny_N= \frac{\left(c^4 \left({x_1}^2+{x_1} {x_2}+{x_2}^2\right)+{x_1}^3 {x_2}^3\right) \left(c^4+{x_1} {x_2} \left({x_1}^2+{x_1} {x_2}+{x_2}^2\right)\right)}{c^2 {x_1}^2 {x_2}^2 ({x_1}+{x_2})^2}. $$ If $x_1$ and $x_2$ are both positive, the above expression attains its minimum $4c^2$ for $x_1=x_2=c$. It follows that $x_Ny_N>c^2$ and $N$ doesn't lie on the hyperbola.
It is then impossible that three normals at different points on the hyperbola meet at the same point on the hyperbola.