Normal to a rectangular hyperbola at P meets the transverse axis at N. Foci of hyperbola are S and S^' , then $6{\left(\frac{SP}{SN}\right)}^2$ is equal to

 We know that PN is the angle bisector of focal radii SP and S'P.
$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{\mathrm{SP}}{{\mathrm{S}}^{\mathrm{\prime }}\mathrm{P}}=\frac{\mathrm{SN}}{{\mathrm{S}}^{\mathrm{\prime }}\mathrm{N}}\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{\mathrm{SN}}{\mathrm{SP}}=\frac{{\mathrm{S}}^{\mathrm{\prime }}\mathrm{N}}{{\mathrm{S}}^{\mathrm{\prime }}\mathrm{P}}=\left|\frac{\mathrm{SN}-{\mathrm{S}}^{\mathrm{\prime }}\mathrm{N}}{\mathrm{SP}-{\mathrm{S}}^{\mathrm{\prime }}\mathrm{P}}\right|=\frac{{\mathrm{SS}}^{\mathrm{\prime }}}{\left|\mathrm{SP}-{\mathrm{S}}^{\mathrm{\prime }}\mathrm{P}\right|}=\mathrm{e}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}6{\left(\frac{\mathrm{SP}}{\mathrm{SN}}\right)}^2=6\cdot \frac{1}{2}=3\end{array}$