The area of the region bounded by hyperbola \(x^2-y^2=9\) and its latus rectum is

\(\begin{aligned}
& x^2-y^2=9 \\
& \Rightarrow \mathrm{a}=\mathrm{b}=1
\end{aligned}\)
\(\Rightarrow\) co-ordinates of latus rectum are
\(\left( \pm a e, \frac{b^2}{a}\right)=( \pm 3 \sqrt{2}, 3)\)

\(\therefore \quad\) Area of hyperbola and its latus rectum
\(=4 \int_3^{3 \sqrt{2}} y \mathrm{~d} x \)
\( =4 \int_3^{3 \sqrt{2}}\left(\sqrt{x^2-9}\right) \mathrm{d} x \)
\( =4\left[\frac{x}{2} \sqrt{x^2-9}-\frac{9}{2} \log \left|x+\sqrt{x^2-9}\right|\right]_3^{3 \sqrt{2}} \)
\(=4[(\frac{3 \sqrt{2}}{2} \sqrt{(3 \sqrt{2})^2-9}-\frac{9}{2} \log |3 \sqrt{2}\) \(+\sqrt{(3 \sqrt{2})^2-9}|)\)
\( \left.-\left(\left.\frac{3}{2} \sqrt{3^2-9}-\frac{9}{2} \log \right\rvert\, 3+\sqrt{3^2-9}\right)\right]\)
\(\begin{aligned} & =4\left[\left(\frac{3 \sqrt{2}}{2} \times 3-\frac{9}{2} \log |3 \sqrt{2}+3|+\frac{9}{2} \log 3\right)\right] \\ & =4\left[\frac{9 \sqrt{2}}{2}-\frac{9}{2} \log (3 \sqrt{2}+3)+\frac{9}{2} \log 3\right] \\ & =4\left[\frac{9 \sqrt{2}}{2}-\frac{9}{2} \log \left(\frac{3 \sqrt{2}+3}{3}\right)\right] \\ & =4\left[\frac{9 \sqrt{2}}{2}-\frac{9}{2} \log (\sqrt{2}+1)\right] \\ & =4 \times \frac{9}{2}[\sqrt{2}-\log (\sqrt{2}+1)] \text { sq. units } \\ & =18[\sqrt{2}-\log (\sqrt{2}+1)] \text { sq. units }\end{aligned}\)