$\begin{aligned} & m_{P K}=m_{Q R} \\ & \frac{2 a t-0}{a t^2-2 a}=\frac{2 a t^{\prime}-2 a r}{a\left(t^{\prime}\right)^2-a r^2} \\ & \frac{t}{t^2-2}=\frac{t^{\prime}-r}{\left(t^{\prime}\right)^2-r^2} \\ & -t^{\prime}-t r^2=-t-r t^2-2 t^{\prime}+2 r, \quad t t^{\prime}=-1 \\ & t^{\prime}-t r^2=-t+2 r-r t^2 \\ & -t r^2+r\left(t^2-2\right)+t^{\prime}+t=0 \\ & \lambda=\frac{\left(2-t^2\right) \pm \sqrt{\left(t^2-2\right)^2+4\left(-1+t^2\right)}}{-2 t} \\ & =\frac{\left(2-t^2\right) \pm \sqrt{t^4}}{-2 t}=\frac{2-t^2 \pm t^2}{-2 t}\end{aligned}$
$r=-\frac{1}{t}$ It is not possible as the $R$ \& $Q$ will be one same.
or $r=\frac{t^2-1}{t}$