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  • Need solution for RD Sharma maths class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 35

Need solution for RD Sharma maths class 12 chapter 20 Areas of Bounded Regions exercise multiple choice question 35

Answers (1)

Answer:

 (b)

Hint:

 Area in all quadrants are equal. So, we do 4 X Area of one quadrant

Given:

 \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1

Explanation:

Area of the ellipse

\begin{aligned} &=4\times \text { Area of OAB }\\ &=4\times \int_{0}^{a}\frac{b}{a}\sqrt{a^{2}-x^{2}}dx \qquad \qquad \qquad \dots (i)\\ &\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\\ &\frac{y^{2}}{b^{2}}=1-\frac{x^{2}}{a^{2}}\\ &\frac{y^{2}}{b^{2}}=\frac{a^{2}-x^{2}}{a^{2}}\\ &y=\frac{b}{a}\sqrt{a^{2}-x^{2}} \end{aligned}

From (i)

\begin{aligned} &=\frac{4b}{a}\int_{0}^{a}\sqrt{a^{2}-x^{2}}dx\\ &=\frac{4b}{a}\left [ \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\frac{x}{a} \right ]_{0}^{a}\\ &=\frac{4b}{a} \left [ \frac{a}{2}\sqrt{a^{2}-a^{2}}+\frac{a^{2}}{2}sin^{-1}\frac{a}{a}-\frac{0}{2}\sqrt{a^{2}-0}-\frac{a^{2}}{2}sin^{-1}\frac{0}{a} \right ]\\ &=\frac{4b}{a} \left [ 0+\frac{a^{2}}{2}sin^{-1}1-0-0 \right ]\\ &=\frac{4b}{a} \left [ \frac{a^{2}}{2}\left ( \frac{\pi }{2} \right ) \right ]\\ &=\pi ab \end{aligned}

Posted by

Gurleen Kaur

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