If ${\mathrm{S}}_1$ and ${\mathrm{S}}_2$ are the foci of the hyperbola whose transverse axis length is 4 and conjugate axis length is 6, ${\mathrm{S}}_3$ and ${\mathrm{S}}_4$ are the foci of the conjugate hyperbola, then the area of the quadrilateral ${\mathrm{S}}_1 {\mathrm{S}}_2 {\mathrm{S}}_3 {\mathrm{S}}_4$ is 
 

Let $\mathrm{H}\equiv \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$

$$\begin{gathered} \mathrm{C}=\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=-1 \\ \mathrm{Given} 2\mathrm{a}=4, 2\mathrm{b}=6 \\ \Rightarrow \mathrm{a}=2, \mathrm{b}=3 \end{gathered}$$

We know that quadrilateral formed by the hyperbola and conjugate hyperbola is a square i.e ${\mathrm{S}}_1{\mathrm{S}}_2={\mathrm{S}}_2 {\mathrm{S}}_3={\mathrm{S}}_3 {\mathrm{S}}_4={\mathrm{S}}_4 {\mathrm{S}}_1$

Required area $={({\mathrm{S}}_1 {\mathrm{S}}_2)}^2$

$$\begin{gathered} ={\mathrm{a}}^2{{\mathrm{e}}_1}^2+{\mathrm{b}}^2{{\mathrm{e}}_2}^2 \\ ={\mathrm{a}}^2\left(\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{{\mathrm{a}}^2}\right)+{\mathrm{b}}^2\left(\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{{\mathrm{b}}^2}\right) \\ =2({\mathrm{a}}^2+{\mathrm{b}}^2) \\ =26 \end{gathered}$$