A tangent to the hyperbola $\frac{{\mathrm{x}}^2}{4}-\frac{{\mathrm{y}}^2}{2}=1$ meets x-axis at P and y-axis at Q. lines $\overline{PR}$and $\overline{QR}$ are drawn such that OPRQ is a rectangle (where O is the origin) then R lies on 

Equation of tangent to hyperbola $\frac{{\mathrm{x}}^2}{4}-\frac{{\mathrm{y}}^2}{2}=1$ is $\frac{\mathrm{x}}{2} \mathrm{sec\theta }-\frac{\mathrm{y}}{\sqrt{2}} \mathrm{tan\theta }=1$

$\Rightarrow \frac{\mathrm{x}}{{\displaystyle \frac{2}{\mathrm{sec\theta }}}}-\frac{\mathrm{y}}{{\displaystyle \frac{\sqrt{2}}{\mathrm{tan\theta }}}}=1$

It meets X-axis at $\mathrm{P}=\left(\frac{2}{\mathrm{sec\theta }}, 0\right)$ and y-axis at $\mathrm{Q}=\left(0, \frac{-\sqrt{2}}{\mathrm{tan\theta }}\right)$

since OPRQ is a rectangle then $\mathrm{R}=\left(\frac{2}{\mathrm{sec\theta }}, \frac{-\sqrt{2}}{\mathrm{tan\theta }}\right)$

let $\mathrm{R}=(\mathrm{x}, \mathrm{y})=\left(\frac{2}{\mathrm{sec\theta }}, \frac{-\sqrt{2}}{\mathrm{tan\theta }}\right)$

$$\begin{gathered} \Rightarrow x=\frac{2}{sec\theta } and y=\frac{-\sqrt{2}}{\tan \theta } \\ \Rightarrow \mathrm{sec\theta }=\frac{2}{\mathrm{x}}, \mathrm{tan\theta }=\frac{-\sqrt{2}}{\mathrm{y}} \\ \mathrm{we} \mathrm{K}.\mathrm{T} {\sec }^2\mathrm{\theta }-{\tan }^2\mathrm{\theta }=1 \\ \Rightarrow \frac{4}{{\mathrm{x}}^2}-\frac{2}{{\mathrm{y}}^2}=1 \end{gathered}$$