The area of the region bounded by the hyperbo | vedclass

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(D) The equation of the hyperbola is $x^2-y^2=9$,which can be written as $\frac{x^2}{3^2}-\frac{y^2}{3^2}=1$.Here,$a=3$ and $b=3$.The eccentricity $e

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$18[\sqrt{2}-\log (\sqrt{2}+1)]$ sq. units

Vedclass · Answer

(D) The equation of the hyperbola is $x^2-y^2=9$,which can be written as $\frac{x^2}{3^2}-\frac{y^2}{3^2}=1$.Here,$a=3$ and $b=3$.The eccentricity $e = \sqrt{1+\frac{b^2}{a^2}} = \sqrt{1+\frac{9}{9}} = \sqrt{2}$.The latus rectum is at $x=ae = 3\sqrt{2}$.The area bounded by the hyperbola and its latus rectum in the first quadrant is $\int_{a}^{ae} y \, dx = \int_{3}^{3\sqrt{2}} \sqrt{x^2-9} \, dx$.Since the hyperbola is symmetric about both axes,the total area is $4 \int_{3}^{3\sqrt{2}} \sqrt{x^2-9} \, dx$.Using the formula $\int \sqrt{x^2-a^2} \, dx = \frac{x}{2}\sqrt{x^2-a^2} - \frac{a^2}{2}\log|x+\sqrt{x^2-a^2}|$,we get:Area $= 4 \left[ \frac{x}{2}\sqrt{x^2-9} - \frac{9}{2}\log|x+\sqrt{x^2-9}| \right]_{3}^{3\sqrt{2}}$.$= 4 \left[ (\frac{3\sqrt{2}}{2}\sqrt{18-9} - \frac{9}{2}\log|3\sqrt{2}+\sqrt{18-9}|) - (0 - \frac{9}{2}\log|3+0|) \right]$.$= 4 \left[ \frac{3\sqrt{2}}{2}(3) - \frac{9}{2}\log(3\sqrt{2}+3) + \frac{9}{2}\log(3) \right]$.$= 4 \left[ \frac{9\sqrt{2}}{2} - \frac{9}{2}\log(\frac{3\sqrt{2}+3}{3}) \right]$.$= 4 \left[ \frac{9\sqrt{2}}{2} - \frac{9}{2}\log(\sqrt{2}+1) \right]$.$= 18[\sqrt{2}-\log(\sqrt{2}+1)]$ sq. units.

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

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