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Provide solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 26

Answers (1)

Answer:

a b\left\{e \sqrt{1-e^{2}}+\sin ^{-1} e\right\}

Hint:

Use ellipse formula

Given:

Find area bounded by ellipse \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 and ordinate x=ae and x=0 whereb^{2}=a^{2}\left(1-e^{2}\right) and e< 1

Solution:

Required area= Area of region

                    = Area ofBORQSP      

                    \begin{aligned} &=2 \times \text { Area OBPS } \\\\ &=2 \int_{0}^{a e} y d x \end{aligned}

We know that,

\begin{aligned} &\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \\\\ &\frac{y^{2}}{b^{2}}=\frac{a^{2}-x^{2}}{a^{2}} \\\\ &y^{2}=\frac{b^{2}}{a^{2}}\left(a^{2}-x^{2}\right) \end{aligned}

\begin{aligned} &y=\pm \sqrt{\frac{b^{2}}{a^{2}}\left(a^{2}-x^{2}\right)} \\\\ &y=\pm \frac{b}{a} \sqrt{a^{2}-x^{2}} \end{aligned}

Since OBPS in first quadrant, value of y is positive

\begin{aligned} &y=\frac{b}{a} \sqrt{a^{2}-x^{2}} \\\\ &=2 \int_{0}^{a e} y d x \end{aligned}

Required area =

\begin{aligned} &=2 \int_{0}^{a e} \frac{b}{a} \sqrt{a^{2}-x^{2}} d x \\\\ \end{aligned}

=\frac{2 b}{a} \int_{0}^{a e} \sqrt{a^{2}-x^{2}} d x \quad\left[\because \int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c\right]

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

\begin{aligned} &=\frac{2 b}{a}\left[\frac{a e}{2} a \sqrt{1-e^{2}}+\frac{a^{2}}{2} \sin ^{-1} e-0\right] \\\\ &=\frac{2 b}{a}\left[\frac{a^{2}}{2} e \sqrt{1-e^{2}}+\frac{a^{2}}{2} \sin ^{-1} e\right] \end{aligned}

\begin{aligned} &=\frac{2 b}{a} \times \frac{a^{2}}{2}\left[e \sqrt{1-e^{2}}+\sin ^{-1} e\right] \\\\ &=a b\left[e \sqrt{1-e^{2}}+\sin ^{-1} e\right] \end{aligned}

 

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