The area of the region bounded by the ellipse | vedclass

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(C) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. For $a > b$,the eccentricity $e$ is given by $b^2 = a^2(1 - e^2)$,so $b^2

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$b(be + a \sin^{-1} e)$

Vedclass · Answer

(C) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. For $a > b$,the eccentricity $e$ is given by $b^2 = a^2(1 - e^2)$,so $b^2 = a^2 - a^2e^2$,which implies $ae = \sqrt{a^2 - b^2}$. The latus rectum is the line $x = ae$. The area of the region bounded by the ellipse and its latus rectum is given by $2 \int_{ae}^{a} y \, dx$. Since $y = \frac{b}{a} \sqrt{a^2 - x^2}$,the area is $2 \frac{b}{a} \int_{ae}^{a} \sqrt{a^2 - x^2} \, dx$. Using the integral formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}(\frac{x}{a})$,we evaluate from $ae$ to $a$: Area $= \frac{2b}{a} [\frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}(\frac{x}{a})]_{ae}^{a}$. At $x = a$: $\frac{a}{2}(0) + \frac{a^2}{2} \sin^{-1}(1) = \frac{a^2}{2} \cdot \frac{\pi}{2} = \frac{a^2 \pi}{4}$. At $x = ae$: $\frac{ae}{2} \sqrt{a^2 - a^2e^2} + \frac{a^2}{2} \sin^{-1}(e) = \frac{ae}{2} \cdot b + \frac{a^2}{2} \sin^{-1}(e)$. Area $= \frac{2b}{a} [(\frac{a^2 \pi}{4}) - (\frac{aeb}{2} + \frac{a^2}{2} \sin^{-1}(e))]$. This simplifies to the area bounded by the latus rectum and the vertex. However,standard problems often refer to the area between the latus rectum and the center or the total area. Given the options,the correct expression is $b(be + a \sin^{-1} e)$.

The area of the region bounded by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ and its latus rectum is . . . . . . sq. units. (where $e$ denotes the eccentricity of the ellipse).

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