The area of the region bounded by the hyperbo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Similar Questions

The area (in square units) of the region bounded by the curves $y + 2x^2 = 0$ and $y + 3x^2 = 1$ is equal to

Find the area common to the curves $y = \sqrt{9 - x^2}$ and $x^2 + y^2 = 6x$.

Find the area enclosed by the parabola $4y = 3x^{2}$ and the line $2y = 3x + 12$.

The area (in sq. units) bounded by the curves $y=\frac{8}{x}$,$y=2x$ and $x=4$ is

The area of the region enclosed by the curves $y = x$,$y = \frac{1}{x}$,$x = e$ and the positive $X$-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module