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

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Aicharya · Accepted Answer

$\begin{aligned} & x^2-y^2=9 \ & \Rightarrow \mathrm{a}=\mathrm{b}=1 \end{aligned}$ $\Rightarrow$ co-ordinates of latus rectum are $\left( \pm a e, \frac{b^2}{a}ight)=( \pm 3 \sqrt{2}, 3)$ $	herefore \quad$ Area of hyperbola and its latus rectum $\begin{aligned} & =4 \int_3^{3 \sqrt{2}} y \mathrm{~d} x \ & =4 \int_3^{3 \sqrt{2}}\left(\sqrt{x^2-9}ight) \mathrm{d} x \ & =4\left[\frac{x}{2} \sqrt{x^2-9}-\frac{9}{2} \log \left|x+\sqrt{x^2-9}ight|ight]_3^{3 \sqrt{2}} \ & =4\left[\left(\frac{3 \sqrt{2}}{2} \sqrt{(3 \sqrt{2})^2-9}-\frac{9}{2} \log \left|3 \sqrt{2}+\sqrt{(3 \sqrt{2})^2-9}ight|ight)ight. \ & \left.-\left(\left.\frac{3}{2} \sqrt{3^2-9}-\frac{9}{2} \log ightvert\, 3+\sqrt{3^2-9}ight)ight] \end{aligned}$ $\begin{aligned} & =4\left[\left(\frac{3 \sqrt{2}}{2} 	imes 3-\frac{9}{2} \log |3 \sqrt{2}+3|+\frac{9}{2} \log 3ight)ight] \ & =4\left[\frac{9 \sqrt{2}}{2}-\frac{9}{2} \log (3 \sqrt{2}+3)+\frac{9}{2} \log 3ight] \ & =4\left[\frac{9 \sqrt{2}}{2}-\frac{9}{2} \log \left(\frac{3 \sqrt{2}+3}{3}ight)ight] \ & =4\left[\frac{9 \sqrt{2}}{2}-\frac{9}{2} \log (\sqrt{2}+1)ight] \ & =4 	imes \frac{9}{2}[\sqrt{2}-\log (\sqrt{2}+1)] 	ext { sq. units } \ & =18[\sqrt{2}-\log (\sqrt{2}+1)] 	ext { sq. units }\end{aligned}$

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

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